# CHAPTER 1 Algebraic Expressions -...

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CHAPTER 1

Algebraic Expressions

1.1 Definition of Terms

1. Algebraic Expression - any collection of symbols, letters, numbers, and operations whichmay represent a value.

Example 1.1.1. The following are algebraic expressions: ,3x2, xy + z

,1

2x3

2x+

7

4.5

x + 1.

2. Constants - numbers or symbols whose values are fixed.

Example 1.1.2. is a constant. The real numbers 4, 6,

3 and 12

are also

constants.

3. Variables - letters or symbols whose values vary.

Example 1.1.3. The letters x and y are commonly used for variables. Usually, it isrepresented by an English or Greek letters.

4. Term - a number, a variable, a product or a quotient of a number and a variable, aproduct of variables, and a quotient of variables.

Example 1.1.4. The terms of 3x5 2x2 + 5 are 3x5,2x2 and 5.

5. Factor - a constant or a variable multiplied together to form a term.

Example 1.1.5. 2, x, x, y are the prime factors of 2x2y.

6. Numerical Coefficient - a factor which is a constant.

7. Literal Coefficient - a factor which is a variable.

Example 1.1.6. 2 is the numerical coefficient of 2x2y, while x2y is the literal coeffi-cient.

8. Like terms or similar terms - terms having the same literal coefficient.

1

9. Unlike terms or dissimilar terms - terms having different literal coefficients.

Example 1.1.7. The terms 4x2y3 and 25x2y3 are like terms while 3xyz and 8x2y3z

are dissimilar terms.

10. Degree of a term - exponent or sum of the exponents of the literal coefficients in theterm.

Example 1.1.8. The degree of 4x3 is 3 since the exponent of x is 3. Moreover, the

degree of1

2x5y4 is 9 since the sum of the exponents of x and y is 5 + 4 = 9.

11. Monomial - an algebraic expression with only one term. It may be a constant, aconstant multiplied to a variable, a variable raised to an exponent, or a product ofconstants and variables raised to an exponent.

12. Binomial - an algebraic expression consisting of two terms.

13. Trinomial - an algebraic expression consisting of three terms.

14. Multinomial - an algebraic expression with more than three terms.

Example 1.1.9. The following are monomials:1

4x3, 12ab2 and

2x

yz.

On the other hand, the expressions 5x + 4y, a3 b3 and 2p 3q5

are binomials, while

the expressions x2 + 4xy + y2,1

3x3 x + 7 and 10 3x2 + x4 are trinomials. An

example of a multinomial is x6 3x3y3 + 8x4y2 2xy5 + 23y6.

15. Polynomial - an algebraic expression whose terms are either a constant, a product ofconstants, or a variable raised to a nonnegative integral exponent.

Example 1.1.10. Observe that 2x2 and1

3x+

7

5are polynomials, but x

13 ,

x+y3z1

and1

x + yare not.

16. Degree of a polynomial - the highest degree among the degrees on the terms in thepolynomial.

2

Example 1.1.11. Given the polynomial 3x3y3 5xy6 + 14x4y, observe that since the

degree of 3x3y3 is 3, the degree of 5xy6 is 7 and the degree of 14x4y is 5, we find

that the degree of the polynomial is 7.

1.2 Operations on Polynomials

1. Addition of PolynomialsAdd the numerical coefficients of liked terms and copy the common literal coefficient.

Example 1.2.1. Perform the indicated operation and simplify if necessary.

1. (3m2 + 4 2m) + (15m2 + 3m 2)

(3m2 + 4 2m) + (15m2 + 3m 2)= (3m2 + 15m2) + (3m 2m) + (4 2)= 18m2 + m + 2

2. (4a2 3a + 10a4) + (2a 3a2 + a3 + 2a4)

(4a2 3a + 10a4) + (2a 3a2 + a3 + 2a4)= (10a4 + 2a4) + (a3) + (4a2 3a2) + (3a + 2a)= 12a4 + a3 + a2 a

2. Subtraction of PolynomialsChange the sign of the terms of the subtrahend and perform addition.

Example 1.2.2. Perform the indicated operation and simplify if necessary.

1. (8a + 2b 3c) (2a 3b + 2c)

(8a + 2b 3c) (2a 3b + 2c) = (8a + 2b 3c) + (2a + 3b 2c)= (8a + (2a)) + (2b + 3b) + (3c 2c)= 6a + 5b 5c

3

2. (2x2 3x + 4) (x2 3x + 5)

(2x2 3x + 4) (x2 3x + 5) = (2x2 3x + 4) + (x2 + 3x 5)= (2x2 + (x2)) + (3x + 3x) + (4 5)= x2 + 0 1= x2 1

3. Multiplication of Polynomials

Definition 1.2.3. If n is a positive integer, we define the nth power of x, denoted byxn, as the product of n factors each equal to x. In symbols, we write

xn = (x)(x) (x) n factors

.

We call x the base, n the exponent, and xn the power.

Laws of ExponentsLet x, y be real numbers and m,n be positive integers.

1. (xm)(xn) = xm+n

2. (xm)n = xmn

3. (xy)m = xmym

Example 1.2.4. Use the laws of exponents to simplify the following expressions.

1. (a4)(a5) = a4+5 = a9

2. (x6)2

= x62 = x123. (st2)

4= (s4)(t2)

4= (s4)(t24) = s4t8

Case 1: Multiplying a monomial to a monomialMultiply the numerical coefficients of each monomial to get the numerical coeffi-cient of the product and multiply the literal coefficients of each monomial to getthe literal coefficient of the product.

4

Example 1.2.5. Perform the indicated operation and simplify if nec-essary.

1. (4a2b3)(3ab5) = (4 3)(a2 a)(b3 b5)= 12a2+1b3+5 = 12a3b8

2. (7x2yz4)(6xy2z3) = (7 6)(x2 x)(y y2)(z4 z3)= 42x2+1y1+2z4+3 = 42x3y3z7

Case 2: Multiplying a polynomial by a monomialMultiply each term of the polynomial by the monomial. Simplify if necesary.

Example 1.2.6. Perform the indicated operation and simplify if nec-essary.

1. 3x2(5x3 2x + 2) = (3x2)(5x3) + (3x2)(2x) + (3x2)(2)= 15x5 6x3 + 6x2

2. 2ab(a2 5ab + 4b2) = (2ab)(a2) + (2ab)(5ab) + (2ab)(4b2)= 2a3b 10a2b2 + 8ab3

Case 3: Multiplying a polynomial by another polynomialMultiply each term of the first polynomial by each term of the second polynomial.Simplify if necesary.

Example 1.2.7. Perform the indicated operation and simplify if necessary.

1. (2x3 x2 + 4x + 1)(2x + 3)

= (2x3)(2x) + (x2)(2x) + (4x)(2x) + (1)(2x) + (2x3)(3) + (x2)(3)+ (4x)(3) + (1)(3)

= 4x4 2x3 + 8x2 + 2x + 6x3 3x2 + 12x + 3= 4x4 + (2x3 + 6x3) + (8x2 3x2) + (2x + 12x) + 3= 4x4 + 4x3 + 5x2 + 14x + 3

2. (3a5 + 2b3)(a2 7ab2 12b5)

= (3a5)(a2) + (3a5)(7ab2) + (3a5)(12b5) + (2b3)(a2)+ (2b3)(7ab2) + (2b3)(12b5)= 3a7 21a6b2 36a5b5 + 2a2b3 14ab5 21b8

4. Division of PolynomialsLaws of ExponentLet x, y be real numbers and m, n be positive integers.

5

1.xm

xn=

xmn if m > n

1 if m = n(1

xmn

)if m < n

2.

(x

y

)m=

xm

ym

Example 1.2.8. Simplify the following using the laws of exponents.

1.x5

x3= x53 = x2

2.(k)4

(k)4= 1

3.d6

d12=

1

d126=

1

d6

4.(w

x

)7=

w7

x7

Case 1: Dividing a monomial by a monomialDivide the numerical coefficients of each monomial to get the numerical coeffi-cient of the quotient and divide the literal coefficients of each monomial to getthe literal coefficient of the quotient.

Example 1.2.9. Perform the indicated operation and simplify if nec-essary.

1. 15a4b3 5a2b = 15a4b3

5a2b=

(15

5

)(a4

a2

)(b3

b

)= 3a42b31 = 3a2b2

2. 8x9y4z2 2xy6z2 == 8x9y4z2

2xy6z2=

(8

2

)(x9

x

)(y4

y6

)(z2

z2

)=

(4)(x91)

(1

y64

)(1) =

4x8

y2

Case 2: Dividing a polynomial by a monomialDivide each term of the polynomial by the monomial. Simplify if necesary.

Example 1.2.10. Perform the indicated operation and simplify if nec-essary.

1. (15a3b3c3 25a2b2c2 + 35abc2) 5abc

=

(15a3b3c3

5abc

)+

(25a2b2c2

5abc

)+

(35abc2

5abc

)= 3a2b2c2 5abc + 7c

6

2. (12a2 7ab + 4b2) ab

=

(12a2

ab

)+

(7abab

)+

(4b2

ab

)=

12a

b 7 + 4b

a

Case 3: Dividing a polynomial by another polynomialSteps:

(a) Arrange the dividend and divisor in descending power of a common literalcoefficient leaving a gap for any missing power of the variable in the dividend.

(b) Divide the first term of the dividend by the first term of the divisor. Thiswill result to the first term of the quotient.

(c) Multiply the divisor by the first term of the quotient and subtract the resultfrom the dividend.

(d) Consider the remainder obtained as the new dividend and repeat (b) and (c)to find the second term of the quotient and the next remainder.

(e) Continue this process until a remainder which is zero or is of lower degree inthe common literal coefficient than the degree of the divisor.

(f) Write the answer in the form

Dividend

divisor= Quotient +

Remainder

divisor

Example 1.2.11. Divide x3 6x2 + 8x + 5 by x 2.

x2 4xx 2 |x3 6x2 + 8x + 5

x3 2x24x2 + 8x + 54x2 + 8x

+5

x3 6x2 + 8x + 5x 2

= x2 4x + 5x 2

7

Case 4: Dividing an nth degree polynomial by a first-degree binomial

Definition 1.2.12. The function defined by the equation P (x) = anxn + an1x

n1 + + a1x + a0, where n is a nonnegative integer and an, an1, . . . , a1, a0 are constants,with an 6= 0 is called a polynomial in x of degree n.

To divide an ndegree polynomial, P (x) by a first-degree binomial, D(x) of the formx r, use the division algorithm known as synthetic division.

Steps in Using Synthetic Division:

(a) Arrange the coefficient of the dividend, P (x) in descending powers of x putting azero for any missing power of x.

(b) Replace the divisor x r by r.(c) Bring down the coefficient of the largest power of x and multiply it by r. Put

the product under the coefficient of the second largest power of x and add thisproduct to the coefficient of the second power of x. Continue this procedure untilthere is a product added to the constant term.

(d) The last number in the third row is the remainder while the rest, reading fromleft to right, are the coefficients of the terms of the quotient, whose degree is oneless than the dividend P (x).

(f) Write the answer in the form

Dividend

divisor= Quotient +

Remainder

divisor

Example 1.2.13. Divide 5x2 14x + 3 by x 2.

2 | 5 14 310 8

5 4 5

5x2 14x + 3x 2

= 5x 4 5x 2

1.3 Special Products

1. Product of Two Binomials: (u + v)(s + t)Use the FOIL method.

8

(a) Multiply the First terms in each binomial.

(b) Multiply the Outer terms in each binomial.

(c) Multiply the Inner terms in each binomial.

(d) Multiply the Last terms in each binomial.

(e) Add the results and simplify if necessary.

Example 1.3.1. Perform the indicated operation and simplify if necessary.

1. (2x + 3y)(3x + 5y)

= (2x)(3x) + (2x)(5y) + (3y)(3x) + (3y)(5y)

= 6x2 + 10xy + 9xy + 15y2

= 6x2 + 19xy + 15y2

2. (ab2 c2d)(2ab + 3cd)

= (ab2)(2ab) + (ab2)(3cd) + (c2d)(2ab) + (cd2)(3cd)= 2a2b3 + 3ab2cd 2abc2d 3c2d3

2. Product of the Sum and Difference of Two Terms: (u + v)(u v)Square the first common term and subtract to it the square of the second common term.

Example 1.3.2. Perform the indicated operation and simplify if necessary.

1. (x + 3y)(x 3y) = (x)2 (3y)2 = x2 9y2

2. (w 5k)(w + 5k) = (w)2 (5k)2 = w2 25k2

3. Square of a Binomial: (u + v)2

Square the first term. Add to it twice the product of the two terms and add the squareof the second term of the binomial.

Example 1.3.3. Perform the indicated operation and simplify if necessary.

1. (x + 3)2 = (x)2 + (2)(x)(3) + (3)2 = x2 + 6x + 9

2. (2y 7)2 = (2y)2 + (2)(2y)(7) + (7)2 = 4y2 28y + 49

4. Cube of a Binomial: (u + v)3

Get the 3rd power of the first term and add to it thrice the product of the square ofthe first term and the second term. Add to the sum thrice the product of the firstterm and the square of the second term. Lastly, add the 3rd power of the second term

9

of the binomial.

Example 1.3.4. Perform the indicated operation and simplify if necessary.

1. (2x + y)3

= (2x)3 + (3)(2x)2(y) + (3)(2x)(y)2 + (y)3

= 8x3 + (3)(4x2)(y) + (3)(2x)(y2) + y3

= 8x3 + 12x2y + 6xy2 + y3

2. (a 3b)2 =

= (a)3 + (3)(a)2(3b) + (3)(a)(3b)2 + (3b)3

= a3 + (3)(a2)(3b) + (3)(a)(9b2) 27b3

= a3 9a2b + 27ab2 27b3

1.4 Factoring

1. Common Monomial Factor

Determine the greatest common factor of all the terms in the given expression, ifthere is any. This would be the common monomial factor.

Divide each term of the expression by the common monomial facotr to determinethe other factor.

Example 1.4.1. Factor completely the given expression.

1. 2x + 10y = 2(x + 5y) 2. 4x2y2 16xy3 = 4xy2(x 4y)

2. Quadratic Trinomial: au2 + bu + cThe factors of a quadratic trinomial consist of two binomials where the first term ofeach binomial is a factor of the first term in the trinomial, the second term of eachbinomial is a factore of the last term in the trinomial. The signs and appropriatefactors to be used are determined by the middle term, which is the sum of the outerand inner products of the two binomial factors. This process is actually the reverse ofthe FOIL method.

10

Example 1.4.2. Factor completely the given expression.

1. x2 8x + 15The factors of 1 are 1 and the factors of 15 are 1,3,5,15. Now sincethe middle term is 8x and the factors whose sum is 8 are 5 and 3, thecomplete factorization of x2 8x + 15 is (x 3)(x 5).

2. 2x2 + 9x 5 = (2x 1)(x + 5)

3. Perfect Square Trinomial: u2 + 2uv + v2

The factors of a perfect square trinomial consist of two like binomials whose terms arethe square roots of the terms which are perfect square and the operation between theterms is determined by the sign of the remaining term in the trinomial.

Example 1.4.3. Factor completely the given expression.

1. y2 + 14y + 49 = (y + 7)2 2. 4b2 4bc + c2 = (2b c)2

4. Difference of Two Squares: u2 v2The factors of the difference of two squares consist of two binomials, one of which isthe sum and the other is the difference of the square roots of the given squares.

Example 1.4.4. Factor completely the given expression.

1. x2 16y2 = (x + 4y)(x 4y) 2. 9k2 25t2 = (3k + 5t)(3k 5t)

5. Sum (Difference) of Two Cubes: u3 v3The factors of the sum (difference) of two cubes consist of a binomial and a trinomial,where the terms of the binomial factor are the cube roots of the given cubic termswith the operation between the terms of the binomial factor the same as that of thegiven binomial, while the terms of the trinomial factor is determined by the binomialfactor, that is, it is consists of the square of the first term in the binomial factor, theproduct of the two terms in the binomial factor and the square of the second termin the binomial factor, with the operation preceding the second term in the trinomialfactor the inverse operation in the given binomial expression.

This rule can be generalized to the sum (difference) of odd powers. Thus, if n is odd,

un vn = (u v)(un1 un2v + un3v2 uvn2 + vn1).

11

Example 1.4.5. Factor completely the given expression.

1. 27x3 8 = (3x 2)[(3x)2 + (3x)(2) + (2)2] = (3x 2)(9x2 + 6x + 4)2. 64p3 +125q3 = (4p+5q)[(4p)2 (4p)(5q)+(5q)2] = (4p+5q)(16p220pq+25q2)3. 32p5 q5 = ((2p)5 + (q)5)

= (2p + (q))((2p)4 (2p)3(q) + (2p)2(q)2 (2p)(q)3 + (q)4

)= (2p q) (16p4 + 8p3q + 4p2q2 + 2pq3 + q4)

4. x7 + 128 = (x7 + (2)7) = (x + 2)(x6 (x5)(2) + (x4)(22) (x3)(23) + (x2)(24)(x)(25) + (26))= (x + 2)(x6 2x5 + 4x4 8x3 + 16x2 32x + 64)

6. Factoring by GroupingThis method is usually applied when the expression to be factored contains four ormore terms.

a. Grouping to Produce a Common FactorGroup terms in the polynomials that would yield a factor that is common to thegroups formed.

Example 1.4.6. Factor completely the given expression.

1. mn + 5m + n + 5We can group this polynomial as (mn+5m)+(n+5) which when simplifiedgives us m(n + 5) + (n + 5). Now by factoring out the common factorbetween the groups, which is (n + 5), we get (n + 5)(m + 1).We can also group this as (mn + n) + (5m + 5). Simplifying each factorwe get n(m + 1) + 5(m + 1). By again factoring out the common factor(m + 1) of the groups, we get (m + 1)(n + 5).

2. a2+ab2b2+2a2b = (a2+ab2b2)+(2a2b) = (ab)(a+2b)+2(ab) =(a b)((a + 2b) + 2) = (a b)(a + 2b + 2)

b. Grouping to Produce a Difference of Two SquaresTo obtain a difference of two squares through grouping, we usually group the termsin the expression that forms a perfect square trinomial since it can be factored asthe square of a binomial.

12

Example 1.4.7. Factor completely the given expression.

1. 4x4 + 4x2y2 + y4 z2 = (4x4 + 4x2y2 + y4) (z2) = (2x2 + y2)2 (z)2 =(2x2 + y2 + z)(2x2 + y2 z)

2. 9a2 b2 + 8bc 16c2 = (3a)2 (b2 8bc + 16c2) = (3a)2 (b 4c)2 =(3a (b 4c))(3a + (b 4c)) = (3a b + 4c)(3a b + 4c)

3. 4w2 + 12wx + 9x2 25y2 + 20yz 4z2 = (4w2 + 12wx + 9x2) (25y2 20yz + 4z2) = (2w + 3x)2 (5y 2z)2 = ((2w + 3x) + (5y 2z))((2w +3x) (5y 2z)) = (2w + 3x + 5y 2z)(2w + 3x 5y + 2z)

7. Adding and Subtracting a Perfect SquareThe goal in this method is also to form a difference of two squares. To do so, a perfectsquare is added to the expression to make it a perfect square trinomial, but is alsosubtracted to preserve the value of the original expression.

Example 1.4.8. Factor completely the given expression.

1. k4 + 64Adding and subtracting 16k2 gives us k4 + 64 + 16k2 16k2. Now by groupingwe get (k4 + 16k2 + 64) 16k2. Simplifying the expression we have,

(k2 + 8)2 (4k)2 = ((k2 + 8) + 4k)((k2 + 8) 4k)

= (k2 + 4k + 8)(k2 4k + 8).

2. 16x4 24x2y2 + 25y4The middle term in this expression should be 40x2y2. Thus we need to add andsubtract 64x2y2 to the expression, which gives us 16x424x2y2+25y4+64x2y264x2y2. Simplifying this we have

(16x4 + 40x2y2 + 25y4) 64x2y2 = (4x2 + 5y2)2 (8xy)2

= ((4x2 + 5y2) + 8xy)((4x2 + 5y2) 8xy)= (4x2 + 5y2 + 8xy)(4x2 + 5y2 8xy).

Remark 1.4.9. Here are some suggestions in factoring polynomials.

Common factors, if any, should be brought out first before applying any other factoringmethod.

If both the difference of two squares and the difference of two cubes are applicable, theformer is done first.

13

If both the sum of two cubes and the sum of odd powers are applicable, the former isused first.

14

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) What do you call the letter used to represent an unknown number?

(a) term (b) constant (c) domain (d) variable

(2) Which of the following expressions is a polynomial?

(a)4

x+ 2 (b)

x

4+ 2 (c) x2 + 4 (d)

x + 4

(3) Which of the following is NOT a monomial?

(a)xy2

z(b) xy2 z (c) 2

7x2y (d) 7

xy

(4) Which of the following expressions is NOT a binomial?

(a) 3x(x + 2) (b)3x + 2

5(c) 4x + 7 (d) (3x + 2)2

(5) Which of the following is a trinomial?

(a) 27x3 (b) 2x2 + 3x 6 (c) 14x5 3x (d) 8x

32x2+x4

(6) The polynomial 2x2y3 + 3xy2 x4y2 is of degree

(a) 3 (b) 6 (c) 5 (d) 8

(7) What is the degree of y3 5y2 + 4y 8?

(a) 1 (b) 3 (c) 5 (d) 6

(8) Which of the following is a set of similar terms?

(a) 7xy, 7x2y2, 7x3y3

(b) 7x2y, 15x2y, 24x2y

(c) 7xy, 7xy2, 7xy3

(d) 7xy2, 15x2y, 24x2y2

15

(9) What is the simplest form of (13m + 2n) (15m 2n)?

(a) 2m (b) 2m 4n (c) 2m (d) 2m + 4n

(10) The sum of (3x 2y) and (2x + 3y) is

(a) x y (b) 5x y (c) 5x + y (d) x + y

(11) The difference when 5x2 + 2x + 7 is subtracted from 10x2 x 8 is

(a) 5x2 + x 1 (b) 5x2 + 3x + 15 (c) 5x2 3x 15 (d) 5x2 +3x+15

(12) The product of x3, 2x2 and 3x is

(a) 6x6 (b) 6x7 (c) 5x6 (d) 5x7

(13) What is the product of (y2 2y + 1) and (y + 3)?

(a) y3+5y2+7y+3 (b) y35y27y+3 (c) y3 +y25y +3 (d) y3y2 +5y3

(14) What is the product of (x + 2) and (2x 3)?

(a) 2x2 x + 6 (b) 2x2 + x 6 (c) 2x2 + 4x 6 (d) 2x2 3x 6

(15) The product of (s + t) and (3s + 2r) is

(a) 3s2 + 2rt

(b) 3s2 + 2rs + 3st + 2rt

(c) 3s2 + 4rst + 3st

(d) 3s2 + 7rs2t2

(16) The quotient when 10x3 + 31x2 + 10x 3 is divided by 5x + 3 is

(a) 2x2 4x + 5 (b) 2x2 + 5x 1 (c) 2x2 3x + 4 (d) 2x2 + 4x + 5

(17) The quotient when 12x2y3 is divided by 3xy is

(a) 36xy2 (b) 4xy2 (c) 15x3y4 (d) 4x3y4

(18) One factor of x2 4x + 4 is x 2. What is the other factor?

(a) x + 2 (b) x 2 (c) x + 4 (d) x 4

16

(19) Which of the following expressions is a perfect square trinomial?

(a) x2 + 3x + 1 (b) x2 + 4x + 4 (c) x2 2x 3 (d) x2 + 5x + 4

(20) The product of the sum of two terms by their difference is

(a) the sum of their squares

(b) the square of their sum

(c) the difference of their squares

(d) the square of their difference

(21) What is the simplest form of (3x 2)2?

(a) 9x2 + 4 (b) 9x2 4 (c) 9x2 6x + 4 (d) 9x2 12x + 4

(22) What is the product of (3x 4)(3x + 4)?

(a) 9x2 16 (b) 9x2 + 16 (c) 9x2 12x + 16 (d) 9x2 24x + 4

(23) The expression (x + y + z)2 when expanded is equal to

(a) x2 + y2 + z2

(b) x2 + 2xy + y2 + 2xz + z2 + 2yz

(c) x2 + y2 + z2 + 2x + 2y + 2z

(d) x2 + y2 z2 + 2xz 2xy + 2yz

(24) What factors form the polynomial 4x2 4x as their product?

(a) 4x and (x 1)(b) (4x 1) and (x + 4)

(c) (2x 1) and (2x + 1)(d) (2x 2) and (2x 2)

(25) What forms the polynomial x2 8x + 16 as its product?

(a) 8(x2 x + 2) (b) (x 4)(x + 4) (c) (x 4)2 (d) (x + 4)2

(26) The common monomial factor of 2x2 + 4x is

(a) 2x (b) 4x (c) 2x2 (d) 4x2

(27) The complete factorization of x3 + 8 is

(a) (x + 2)3

(b) (x + 2)(x2 2x + 4)(c) (x + 2)(x2 + 2x + 4)

(d) (x2 + 2)(x + 4)

17

(28) x6 y6 when completely factored is equal to

(a) (x+y)(xy)(x2xy+y2)(x2+xy+y2)(b) (x3 y3)2

(c) (x4 y4)(x2 + y2)(d) (x + y)2(x2 xy + y2)(x2 + xy + y2)

(29) What are the factors of 2x2 x 1?

(a) (2x 1)(x + 1) (b) (2x + 1)(x 1) (c) (2x)(x 1) (d) (x 2)(x 1)

(30) What is the complete factorization of 4x2 16?

(a) 2(x 4)(x + 4)(b) 4(x 2)(x + 2)

(c) 4(x 4)(d) 2(x 2)(x + 2)

(31) The prime factors of 64x3y2 are

(a) 16 4 x3 y2 (b) 42 22 x3 y2 (c) 26 x3 y2 (d) 43 x3 y2

(32) What are the factors of x2 4x + 3?

(a) (x 3)(x + 1) (b) (x + 3)(x 1) (c) (x 3)(x 1) (d) (x + 3)(x + 1)

(33) Which of the following expressions is NOT factorable?

(a) 14x2 + 7x 21 (b) 8x3 + y3 (c) x4 + 4x (d) 2x + y2

(34) The factors of xy + 2x + y + 2 is

(a) x and (y + 2)

(b) (x + 1) and (y + 2)

(c) x, (x + 1) and (y + 2)

(d) x, (x + 2) and (y + 1)

(35) Which of the following is NOT a factor of x8 y8 ?

(a) x + y (b) x y (c) x2 + y2 (d) x3 + y3

18

CHAPTER 2

Rational Expressions

Definition 2.0.10. A rational expression is an expression that can be written in the formP

Qwhere P and Q are polynomials, Q 6= 0.

2.1 Simplifying Rational Expressions

A rational expression is said to be in its simplest form if the greatest common factor of thenumerator and the denominator is 1.To simplify a rational expression

1. Find the greatest common factor between the numerator and the denominator.

2. Factor out the GCF from the numerator and from the denominator.

3. Divide the common factor/s in the fraction to get the simplified form of the expression.

Example 2.1.1. Reduce the following fractions into its simplest form.

1.24ab

14bc=

(2b)(12a)

(2b)(7c)=

12a

7c2.

x2 y2

x2 + 2xy + y2=

(x + y)(x y)(x + y)(x + y)

=

(x y)(x + y)

Remark 2.1.2. Note that

(i)a

b=ab

= (

a

b

)=

(ab

)

(ii) ab

= (ab

)=

a

b=ab

2.2 Operations on Rational Expressions

The operations on rational expressions can be done using the same methods applied onrational numbers or fractions.

1. Addition (Subtraction) of Rational Expressions

Case 1: Similar Rational ExpressionsAdd (Subtract) their numerators to get the resulting numerator and copy thecommon denominator.

19

Case 2: Dissimilar Rational ExpressionsConvert the given expressions into similar fractions before performing the indi-cated operation.

2. Multiplication of Rational ExpressionsThe product of two rational expressions is a rational expression whose numerator isthe product of the numerators of the given expressions and whose denominator is theproduct of the denominators of the given expressions.

3. Division of Rational ExpressionsDividing rational expressions is the same as multiplying the dividend by the reciprocalof the divisor.

Recall that the reciprocal of a rational expressionN

Dis

D

N.

Example 2.2.1. Perform the indicated operations and simplify the final answer.

1.m

5(

2m 105

)=

m (2m 10)5

=m + (2m + 10)

5=m + 10

5=

10m5

2.

(x2 4

3x

)+

(5x2

x 2

)=

((x2 4)(x 2)

(3x)(x 2)

)+

((5x2)(3x)

(3x)(x 2)

)

=(x2 4)(x 2) + (5x2)(3x)

(3x)(x 2)

=(x2)(x) + (x2)(2) + (4)(x) + (4)(2) + (5x2)(3x)

(3x)(x 2)

=x3 2x2 4x + 8 + 15x3

(3x)(x 2)

=16x3 2x2 4x + 8

(3x)(x 2)

3.12ck2

4k(

6c2k

13

)=

12ck2

4k(

13

6c2k

)=

(12ck2)(13)

(4k)(6c2k)=

156ck2

24c2k2=

(12ck2)(13)

(12ck2)(2c)=

13

2c

20

4.(x + 1)

(x 1)(x + 2) 6(x + 1)

5(x 1)=

(x + 1)(6)(x + 1)

(x 1)(x + 2)(5)(x 1)

=(6)(x2 + 2x + 1)

(5)(x + 2)(x2 2x + 1)

=6x2 + 12x + 6

(5x + 10)(x2 2x + 1)

=6x2 + 12x + 6

5x3 10x2 + 5x + 10x2 20x + 10

=6x2 + 12x + 6

5x3 15x + 10

2.3 Complex Fractions

Definition 2.3.1. A complex fraction is a fraction which has a fraction in its numeratoror in its denominator or in both.

To simplify a complex fraction, perform all the indicated operations in both the numeratorand in the denominator.

Example 2.3.2. Simplify the following complex fractions.

1.

y x2

y

y2

x x

=

y2 x2

y

y2 x2

x

=

(y2 x2

y

)(

y2 x2

x

)=

(y2 x2

y

)(

x

y2 x2

)=

x

y

2.k

1 1

1 +1

k 1

=k

1 1((k 1) + 1

k 1

) = k1 1(k

k 1

)

=k

1(

1(

k

k 1

)) = k1

(1

(k 1

k

)) = k1

(k 1

k

)=

k

k (k 1)k

=k

k k + 1k

=k(1

k

) = k (1k

)= k k = k2

21

2.4 Integral Exponents Leading to Complex Fractions

Definition 2.4.1. For any positive integer n and for any nonzero real number x, y

1. x0 = 1 2. xn =1

xn 3.

(x

y

)n=(y

x

)nNote: Algebraic expressions involving integral exponents leads to complex fractions.

Example 2.4.2. Simplify the following expressions.

1. a0b2c4 = (1)

(1

b2

)(c4) =

c4

b2

2.

(m3

n2

)5=

(n2

m3

)5=

n25

m35=

n10

m15=

n10

1

m15

= n10 (

1

m15

)= n10m15

3.a1 + b1

(a + b)1=

1

a+

1

b1

(a + b)

=

b + a

ab1

(a + b)

=

(b + a

ab

)(

1

a + b

)=

(b + a

ab

)(a + b

1

)=

(a + b)2

ab

22

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) Which of the following is NOT equal to3

7?

(a)37

(b) (37

) (c) (37

)(d)

(3

7

)

(2) The simplified form of6x 6y3x 3y

is

(a) 0 (b) 2 (c) 2(x y) (d) x y

(3) The simplified form of

(2x2y

z

)3is

(a)2x6y3

z3(b)

2x5y3

z3(c)

6x6y3

z3(d)

8x6y3

z3

(4)x2 + x

xwhen reduced to lowest terms is equal to

(a) x (b) x 1 (c) x + 1 (d) x2

(5)12y2 + y 16y2 y 1

is equal to

(a)4y 12y 1

(b)2y 1y 1

(c)2y + 1

y + 1(d)

4y + 1

2y + 1

(6) The least common multiple of 9a2 1 and 9a + 3 is

(a) 9a2 + 3

(b) (3a + 1)(3a 1)(c) 3(3a + 1)(3a 1)(d) 3(3a 1)

(7) The sum of1

x+

2

x2+

3

x3is

(a)x3 + 2x2 + 3x

x3(b)

6

x + x2 + x3(c)

6

x3(d)

3 + 2x + x2

x3

(8) The LCD of6xy

9x2y,

4y

18x3y2and

2

3xyis

23

(a) 3xy (b) 18x2y (c) 9x2y (d) 18x3y2

(9) What is the LCD of4x y

3xand

2y + 2

6xy?

(a) 6xy (b) 18xy (c) 3xy (d) 6x2y

(10) The expression2x

3x 1+

3x

3x + 1 12x

2

9x2 1is equal to

(a)x

3x + 1 (b)3x2 5x

(3x 1)(3x + 1)(c)

x

3x 1 (d)7x2 3x

(3x + 1)(3x 1)

(11) The difference when2

8 xis subtracted from

3

x 8is

(a)5

x 8

(b)1

8 x

(c)1

x + 8

(d)5

8 x

(12) The simplified form of

(a4

2b2

)(2b3

a

)2is

(a) a9b2 (b) a6b2 (c) 2a2b4 (d) 2a2b7

(13)7ab2

6bc3 18ac28a3b

8b2c

9ais equal to

(a)2b2c

a(b)

27b2c

32a(c)

27

32b2c3(d)

27b2c3

32a

(14) The quotient ofx2y3

5x5z7 2y

5

35x3z8is

(a)7z

2y2(b)

2y2

7z(c)

2z

7y2(d)

7y2

2z

(15) The fraction1 1

3

4 +2

3

is equal to

(a)1

8(b)

1

7(c) 7 (d)

8

9

(16) Which is the simplified form of

y x2

y

y2

x x

?

24

(a)x2

y2(b)

y

x (c)y2

x2(d) x

y

(17) The simplest form of the expression3x2y3

6x4y2is

(a)x6

2y(b)

x6

2y5(c)

2y

x6(d)

2y5

x6

(18) Which of the following is equal to (x1 + y1)1

?

(a) x + y (b)1

x + y(c)

xy

x + y(d)

x + y

xy

(19) (x + y)1(x2 y2) is equal to

(a) y x (b) x y (c) y x(xy)2

(d)1

x2y2

(20) Which expression is equivalent to 2a1 + 3b1?

(a)1

2a + 3b(b)

2a + 3b

ab(c)

12a + 3b

(d)2b + 3a

ab

25

CHAPTER 3

Radicals and Rational Exponents

3.1 Basic Terms and Properties

Definition 3.1.1. If x, y are real numbers and n is a positive integer (n 2) such thatxn = y, then y is called an nth root of x.The nth root of x is not unique, thus we define the principal nth root of x as

n

x =

{the positive nth root of x if x > 0the negative nth root of x if x < 0

n

x is called a radical, where x is the radicand, and n is the order or index. The indexis usually omitted if n = 2.

Definition 3.1.2. For any positive integer n and any real number x, if n

x is a real numberthen

x1n = n

x.

In general, for any positive integers m,n which are relatively prime, and any real number xfor which n

x is a real number

xmn =

( n

x)m

orn

(xm)

Laws on RadicalsLet x, y be real numbers and m,n be positive integers such that n

x, n

y, and mn

x are realnumbers.

1. ( n

x)n = x

2. n

xy = n

x n

y

3. n

x

y=

n

xn

y

4. m

n

x = mn

x

3.2 Simplifying Radicals

A radical is in its simplest form if it satisfies the following conditions:

1. The radicand contains no factor that is a power having an exponent greater than orequal to the index.If not, extract all possible roots of the factors in the radicand.

26

2. The index cannot be reduced any further.If not, express the radical using rational exponents and reduce the rational exponentto lowest term before expressing it back to radical form.

3. The radicand contains no fraction.If not, rationalize the denominator of the fraction in the radicand.To rationalize, multiply the both the numerator and the denominator of the radicalby a factor (which is also a radical) that would yield a product with no radicals in thedenominator.

4. There is no radical in the denominator of a fraction.If not, rationalize the radical in the denominator of the fraction.

Example 3.2.1. Simplify the following expressions.

1. 823 =

3

82 = 3

(23)2 = 3

(22)3 = 22 = 4

2. 8134 =

4

813 =4

(1

81

)3=

4

((12

)4)3=

4

((12

)3)4=

(1

2

)3=

1

8

3.

200 =

100 2 =

100

2 = 10

2

4. 6

49 =6

72 = 726 = 7

13 = 3

7

5.

5

7=

57

77

=

5

7

7=

35

7

6.4

3

2=

4

3

2 3 +

2

3 +

2=

4(3 +

2)

32 (

2)2=

12 + 4

2

9 2=

12 + 4

2

7

7.4

25x4 = 4

(5x2)2 = (5x2)24 = (5x2)

12 =

5x2 =

5

x2 = x

5

8.3

m5

n=

3

m5

3

n=

3

m5

3

n

3

n2

3

n2=

3

m53

n2

3

n3=

3

m5n2

n=

3

m33

m2n2

n=

m3

m2n2

n

9.5x3

9=

5x3

9

3

33

3=

5x 3

33

27=

5x 3

33

33=

5x 3

3

3

27

3.3 Operations on Radicals

1. Addition (Subtraction) of RadicalsRadicals like algebraic expressions can be added (subtracted) if they are similar.

Definition 3.3.1. Two radicals are similar if they have the same index and the sameradicand.

Example 3.3.2. 4 3

x and 3w 3

x are similar radicals.

To add (subtract) two radicals, add (subtract) the coefficients outside the radical andcopy the common radical. Simplify if necessary.

Example 3.3.3. Perform the indicated operations and simplify if necessary.

1. 3

2 + 2

18

2

3

2 + 2

18

2 = 3

2 + 2

9 2

2

= 3

2 + 2

32 2

2

= 3

2 + 2

32

2

2

= 3

2 + 2 3

2

2

= 3

2 + 6

2

2

= (3 + 6 2)

2

= 7

2

2. 4

64 + 2

32

4

64 + 2

32 =4

26 + 2

16 2= 2

64 + 2

42 2

= 232 + 2

42

2

=

23 + (2 4)

2

=

22 2 + 8

2

=

22

2 + 8

2

= 2

2 + 8

2

= 10

2

28

3.3

27x4 + 364x7 + 7 3

x

3

27x4 +364x7 + 7 3

x =

3

33x3x + 3

(4)3x6x + 7 3

x

=3

333

x3 3

x + 3

(4)3 3

x6 3

x + 7 3

x

= 3x 3

x + (4x2) 3

x + 7 3

x

= (3x + (4x2) + 7) 3

x

= (4x2 + 3x + 7) 3

x

2. Multiplication of Radicals

Case 1: Radicals of the Same OrderMultiply the coefficients of each expression to get the coefficient of the productand multiply the radicands of each expression to get the radicand of the productand copy the common order or index.

Example 3.3.4. Perform the indicated operations and simplify if necessary.

1. 3

2 3

2x 3

3y = 3

(2)(2x)(3y) = 3

12xy

2.

r

2r3

5r5 =

(r)(2r3)(5r5) =

10r9 =

10r8r

=

10r

r8 = r4

10r

3. 4

36x2y3 4

72x3y

4

36x2y3 4

72x3y = 4

(22 32)x2 y3 4

(23 32)x3 y= 4

22 23 32 32 x2 x3 y3 y

= 4

25 34 x5 y4

= 4

24 2 34 x4 x y4

= 4

24 34 x4 y4 4

2x

= 2 3x y 4

2x

= 6xy4

2x

Case 2: Radicals of Different OrdersConvert the radicals into their equivalent radicals such that they all have the sameindices and apply the rule for Case 1.

29

Example 3.3.5. Perform the indicated operations and simplify if necessary.

1.

x 3

x 4

x= x

12 x 13 x 14 = x 612 x 412 x 312

=12

x6 12

x4 12

x3 =12

x6 x4 x3=

12

x13 =12

x12x =12

x12 12

x= x 12

x

2.3

3x2

2x = (3x2)13 (2x) 12 = (3x2) 26 (2x) 36

= 6

(3x2)2 6

(2x)3 = 6

(32) (x2)2 6

(23) (x3)= 6

(32) (x2)2 (23) (x3) = 6

9 x4 8 x3 = 6

72x7

=6

72x6x =6

x6 6

72x= x 6

72x

3. Division of RadicalsWrite the expression in fractional form and rationalize the denominator.

Example 3.3.6. Perform the indicated operations and simplify if necessary.

1. 36

x6 8

x2 =36

x6

8

x2=

(36

8

)(

x6

x2

)=

(9

2

)(x62

)=

(9

2

)(x4)

=9x4

2

2. (3

6a3 5

4a)

2a

(3

6a3 5

4a)

2a =(3

6a3 5

4a)2a

=(3

6a3 5

4a)2a

2aa

=(3

6a3 5

4a)

2a4a2

=(3

6a3)(

2a) (5

4a)(

2a)

2a

=3

(6a3)(2a) 5

(4a)(2a)

2a

=3

12a4 5

8a2

2a=

3

3 4 a4 5

4 2 a22a

=3

4a4

3 5

4a2

2

2a=

3 (2a2)

3 5 (2a)

2

2a

=2a(3a

3 5

2)

2a= 3a

3 5

2

30

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) Another way of writing (x + y)25 is

(a) 5

(x + y)2 (b)

(x + y)5 (c) x25 + y

25 (d) 5

x2 + y2

(2) (9

32

)is equal to

(a) 27 (b) 3 (c) 18 (d) 81

(3) The product of x14 and x

18 is

(a) x112 (b) x

16 (c) x

132 (d) x

38

(4) The expression

2xy

4

4x2y2when simplified is equal to

(a)

2xy

2xy(b)

2xy4x2y2

(c)1

2(d) 1

(5) When simplified, the expression

(x2y6

9

) 12

is

(a)3y

x3(b)

3x

y3(c)

x

3y3 (d)xy3

3

(6) Which of the following expressions is NOT equal to 1?

(a) (3x2)0

(b)

1(c)

3x + 44 + 3x

(d)1

(7) The product x23 y

56 can be written in radical form as

(a) 3

x2y5 (b) 6

x4y5 (c) y 3

x2y2 (d) 6

x2y5

31

(8) What is the simplified form of

x

y?

(a)

x

y(b)

xy

y(c)

xy

y(d)

x

y

(9) When simplified,2

2 +

2is equal to

(a)2

2

2

(b) 2 +

2 (c) 2

2(d)

2 +

2

2

(10) All of the following are equal to x12 y

34 EXCEPT

(a)

x 4

y3 (b) (y3x2)14 (c) 4

x2y3 (d)

(x

14 y

12

) 14

(11) The simplified form of

48a4b3c is

(a)

3abc (b) 16a2b

3bc (c) 4a2b

3bc (d) 4a

3b3c

(12) The sum of

18,

50 and

32 is

(a) 4

2 (b) 4 (c) 12

2 (d) 6

2

(13) The simplified form of

20x3y + x

45xy

5x3y

(a) 4x

5xy (b) 12x

5xy (c) 10x

5xy (d) x

5xy

(14) The simplest form of

2r5t3

r5t6 is

(a) r2t

2rt r2t3

r

(b) (r2t r2t3)(

2rt

r) (c) 2r2trt r2t3t

(d) r4t4

2rt

(15)

8

3

8 6

8 is equal to

(a) 0 (b)

2 (c) 1 (d)

8

32

(16) The product of 9a12 and (8a)

23 is equal to

(a) 72a13 (b) 36a

76 (c) 12a

76 (d) 72a

76

(17) The product of

18x2y and

2xy3, in simplest form, is

(a) 36xy2

x (b) 9xy2

x (c) 2xy2

5x (d) 6xy2

x

(18) The product(

x y 3) (

x y + 3)

is equal to

(a) x y 3(b) (x y)2 9

(c) (x y)2 3(d) x y 9

(19) The radical expression2

3 3

2

4

3 + 2

2when rationalized is

(a)9 + 4

6

10(b)

9 + 4

6

10(c)

9 4

6

10

(d) 2 + 3

6

(20) As a single radical, 3

3

2 can be written as

(a) 5

6 (b) 6

6 (c) 6

36 (d) 6

72

33

CHAPTER 4

Linear and Quadratic Equations and Inequalities

4.1 Equations

Definition 4.1.1. An equation is a statement that shows two quantities or expressions areequal. An equation may be true only for some values of the variable/s.The value of the variable that makes a statement true is called solution or root of theequation.

To solve an equation is to determine its solution/s.Properties of EqualityIf a and b represent the same quantity, then a = b. (read a equals b or a is equal to b.)

1. Substitution Property of EqualityIf a, b are real numbers and a = b, then a may be replaced by b or b may be replacedby a in any statement without changing the meaning of the statement.

2. Reflexive Property of EqualityIf a is real number, then a = a.

3. Symmetric Property of EqualityIf a, b are real numbers and a = b, then b = a.

4. Transitive Property of EqualityIf a, b, c are real numbers, a = b, and b = c, then a = c.

5. Addition Property of Equality (APE)If a, b, c are real numbers and a = b, then a + c = b + c.

6. Multiplication Property of Equality (MPE)If a, b, c are real numbers and a = b, then ac = bc.

7. Property of Zero ProductIf a, b are real numbers and ab = 0, then either a = 0 or b = 0.

8. Square Root Propertyx2 = c is equivalent to x =

c.

34

4.2 Linear Equations in One Variable

Definition 4.2.1. A linear equation in one variable is an first-degree equation that canbe written in the form ax + b = 0 where a, b are real numbers and a 6= 0.

To solve a linear equation in one variable, apply the Addition Property of Equality (APE)and the Multiplication Property of Equality (MPE).

Example 4.2.2. Solve for the value of x.

1. x + 7 = 18

x + 7 = 18

x = 18 7x = 11

2. 3x 5 = 2x 4

3x 5 = 2x 43x 2x = 4 + 5

x = 1

4.3 Linear Equations Involving Absolute Values

Definition 4.3.1. If a is a real number, then the absolute value of a, denoted by |a| isgiven by,

|a| ={

a if a 0a if a < 0

We consider three cases in solving linear equations involving absolute values.

Case 1: Equations of the form |u| = k where u is a linear expression in one variable and k isa constant.If k 0, then the equation |u| = k is equivalent to u = k and u = k.If k < 0,then |u| = k has no solution.

Example 4.3.2. Solve the following equations.

1. |5x 4| = 0

|5x 4| = 0 5x 4 = 0 5x = 4 x = 4

5

35

2. |2x 3| = 5

|2x 3| = 5 2x 3 = 5 or 2x 3 = 5 2x = 5 + 3 2x = 5 + 3 2x = 8 2x = 2 x = 4 x = 1

3. |4x + 3| = 7Since k = 7 < 0 the equation has no solution.

Case 2: Equations of the form |u| = |v| where u and v are linear expression in one variable.|u| = |v| is equivalent to u = v or u = v.

Example 4.3.3. Solve the following equations.

1. |3 x| = |x + 2|

|3 x| = |x + 2| 3 x = (x + 2) or 3 x = (x + 2) 3 x = x + 2 3 x = x 2 3 2 = x + x 3 + 2 = x + x 1 = 2x 5 6= 0 x = 1

2no solution

2. |4x 5| = |1 2x|

|4x 5| = |1 2x| 4x 5 = 1 2x 4x 5 = 1 + 2x 4x + 2x = 1 + 5 4x 2x = 1 + 5 6x = 6 2x = 4 x = 1 and x = 2

Case 3: Equations of the form |u| = v where u and v are linear expression in one variable.The solutions of |u| = v are solutions of either u = v or u = v or both. To identifywhich is/are the solution/s, we substitute the obtained solutions for |u| = |v| to theequation |u| = v. The value/s which will satisfy |u| = v are the solutions to the givenequation.

36

Example 4.3.4. Find the solution/s of

x 45 = x4 .x 45

= x4 x 45 = (x4) or x 45 = (x4) x 4

5=

x

4

x 45

=x4

20(

x 45

)= 20

(x4

)20

(x 4

5

)= 20

(x4

) 4(x 4) = 5(x) 4(x 4) = 5(x) 4x 16 = 5x 4x 16 = 5x 4x 5x = 16 4x + 5x = 16 x = 16 9x = 16 x = 16 x = 16

9

if x = 16 if x = 169

16 45

= 16416

9 4

5

=16

94

205

= 16416 36

95

=16

9 14

|4| = 4

2095

=4

9

4 6= 4209 15

= 49 x = 16 is not a solution

49 = 49

49

=

4

9

x = 49

is the solution

37

Example 4.3.5. Find the solution/s of |3x 4| = x + 1.

|3x 4| = x + 1 3x 4 = (x + 1) or 3x 4 = (x + 1) 3x 4 = x + 1 3x 4 = x 1 3x x = 1 + 4 3x + x = 1 + 4 2x = 5 4x = 3 x = 5

2x =

3

4

if x =5

2if x =

3

4

3(52

) 4 = 52 + 1

3(34) 4 = 34 + 1

152 4

= 52 + 194 4

= 34 + 1

15 82 = 5 + 22

9 164 = 3 + 44

72 = 72

74 = 74

72

=

7

2

7

4=

7

4

4.4 Quadratic Equations in One Variable

Definition 4.4.1. A quadratic equation in one variable is a second-degree equationthat can be written in the form ax2 + bx + c = 0 where a, b, c are real numbers and a 6= 0.

Methods of Solving Quadratic Equations in One Variable

1. FactoringConsider a quadratic equation in standard form with the left member being factorable.Factor the left-hand side of the equation and apply the Property of Zero Product tofind the solution of the equation.

2. Completing the SquaresConsider the quadratic equation ax2 + bx + c = 0, where a > 0.

(a) If a = 1, write the equation in the form x2 + bx = c.If a 6= 1, divide both sides of the equation by a and write the resulting equationin the form x2 + bx = c.

(b) From the equation in (a), take half of the coefficient of x, square it and add theresult to both sides of the equation. The left member of the resulting equationwill be a perfect square trinomial.

38

(c) Write the left-side of the equation in factored form and apply the Square RootProperty.

3. Quadratic FormulaThe roots of the quadratic equation ax2 + bx + c = 0, where a 6= 0 can be obtainedusing the formula

x =b

b2 4ac

2a

Example 4.4.2. Given x2 + x 20 = 0. Solve for the value/s of x using all themethods described above. Solution:

Using Factoring:Observe that the left-hand side of the equation can be factored as (x+5)(x4).Thus we have (x + 5)(x 4) = 0. By the Property of Zero Product, we havex + 5 = 0 and x 4 = 0. Hence we find that x = 5 and x = 4. Thus, thesolution set S = {5, 4}.

Using Completing the Square:Since the coefficienct of x2 in the given equation is 1, we write the equation in theform x2 + bx = c. Thus we have x2 + x = 20. Now we add to both sides of the

equation the value of

(b

2

)2=

(1

2

)2=

1

4. Hence we have x2+x+

1

4= 20+

1

4(

x +1

2

)2=

(20)(4) + 1

4

(x +

1

2

)2=

81

4

(x +

1

2

)2=

(9

2

)2. Now

by applying the square root property we get x+1

2= 9

2. Solving for the values

of x we have x = 12 9

2, which gives us x = 4 and x = 5. Thus, the solution

set is again S = {5, 4}.

Using the Quadratic Formula:Observe that in the given equation a = 1, b = 1 and c = 20. Thus x =(1)

(1)2 4(1)(20)2(1)

=1

1 + 80

2=

1

81

2=

1 92

. This

results to x = 5 and x = 4. Thus the solution set is also S = {5, 4}.

Remark 4.4.3. In the quadratic formula, the radicand D = b2 4ac, called the discrim-inant, describes the nature of the roots of the quadratic equation even if the roots are notknown.If D > 0 the roots are real and unequal.

39

If D = 0 the roots are real and equal.If D < 0 the roots are nonreal(complex) and unequal.

Example 4.4.4. Determine the nature of the roots of the following quadratic equa-tions.

1. 4x2 3x 2 = 0Observe that in the given equation a = 4, b = 3 and c = 2. Thus D =b2 4ac = (3)2 (4)(4)(2) = 9 + 32 = 41. Now since D = 41 > 0, the rootsof 4x2 3x 2 = 0 are real and unequal.

2. 8x2 12x + 5 = 0Note that a = 8, b = 12 and c = 5. Thus D = b2 4ac = (12)2 (4)(8)(5) =144 160 = 16. Since D = 16 < 0, the roots of 4x2 3x 2 = 0 are nonreal(complex) and unequal.

3. 4x2 12x 9 = 0 Observe that a = 4, b = 12 and c = 9. Hence D =b2 4ac = (12)2 (4)(4)(9) = 144 144 = 0. Therefore since D = 0, theroots of 4x2 3x 2 = 0 are real (complex) and equal.

Remark 4.4.5. If the roots r1 and r2 of a quadratic equation are given, the quadraticequation is given by

x2 (r1 + r2)x + r1r2 = 0.Therefore, given a quadratic equation of the form ax2 + bx + c = 0 and its roots r1 and r2,observe that the sum of the roots,

r1 + r2 = b

aand the product of the roots,

r1r2 =c

a.

Example 4.4.6. Find a quadratic equation with the following roots.

1.2

5;2

Solution:

Let r1 =2

5and r2 = 2. Then r1 + r2 =

2

5+ (2) = 2 + (10)

5= 8

5.

Moreover, r1r2 =

(2

5

)(2) = (2)(2)

5=45

.

Thus x2 (85

)x +

(45

)= 0 x2 + 8

5x 4

5= 0.

Note that the quadratic equations with these roots is not unique. Anotherequation with these roots is 5x2 + 8x 4 = 0.

40

2. 3; 5Solution:Let r1 = 3 and r2 = 5. Then r1 + r2 = 3 + 5 = 8. Moreover, r1r2 = (3)(5) = 15.Thus x2 (3)x + 5 = 0 x2 3x + 5 = 0.

3. 34;1

2Solution:

Let r1 = 3

4and r2 =

1

2. Then r1 + r2 =

(3

4

)+

(1

2

)=

(3) + (2)4

=

54.

Moreover, r1r2 =

(3

4

)(1

2

)=

(3)(1)(4)(2)

=3

8.

Thus x2 (5

4

)x +

(3

8

)= 0 x2 + 5

4x +

3

8= 0.

We can also write 8x2 + 10x + 3 = 0 as the equation with these roots.

4.5 Applications of Linear and Quadratic Equations

A word problem describes a situation involving both known and unknown quantities. Manyreal life probles can be solved by expressing them as word problems that often lead toequations.The equation that represents a word problem is called a mathematical model. The success inworking out word problem depends largely on ones ability to translate into a mathematicalmodel the word problem.Steps in Solving a Word Problem

(1) Read the problem carefully and make sure that the situation is thoroughly understood.

(2) Identify the quantities, both known and unknown that are involved in the problem.

(3) Select one of the unknown quantities and represent it by a variable and then expressany other unknown in terms of this variable, if possible.

(4) Search the problem for information that tells what quantities or combination of themare equal. Often, making a sketch helps to carry out this step.

(5) Write an equation using the expressions formed in (4). Carrying through the calcula-tions with an initial guess sometimes helps to clarify the relationship between variables.

(6) Solve the equation obtained in (5).

41

(7) Check the solution in the original problem. This step is critical since the main goal isto find a solution of the stated problem rather than of the equation that was written.

(8) Write the answer to the original problem.

Solve the following problems using linear equations:

Example 4.5.1. A 30-year old father has a 4-year old son. In how many years willthe father be thrice as old as his son?Solution:We want to determine the number of years the fathers age will be thrice the sonsage. Thus we let x be the number of years the fathers age will be thrice the sonsage. Now observe that the following information can be obtained from the problem:

Present Age Age after x yearsFather 30 30 + xSon 4 4 + x

.

Thus we have 30 + x = 3(4 + x). Hence solving for x we find that x = 9. Thus, thefathers age will be thrice the sons age after 9 years.

Example 4.5.2. The sum of the digits of a two-digit number is 12. The value ofthe number is 6 less than 9 times the units digit. What is the number?Solution:Let x = units digit; 12 x = tens digit.

10(12 x) + x = 9x 6120 10x + x = 9x 6

120 + 6 = 9x + 9x

126 = 18x

x = 7

12 x = 12 7 = 5

The number is 57.

42

Example 4.5.3. How many liters of a 70% alcohol solution should be added to 200liters of a 20% alcohol solution to make it 35% pure?Solution:

Alcohol (in %) Alcohol Solution (in Liters) Alcohol (in Liters)70% alcohol x 0.7x20% alcohol 200 (0.2)(200)35% alcohol 200 + x (0.35)(200 + x)

.

0.7x + (0.2)(200) = (0.35)(200 + x)

0.7x + 40 = 70 + 0.35x

0.7x 0.35x = 70 400.35x = 30

x = 85.7143

85.7143 liters of 70% alcohol solution must be added to 200 liters of 20% alcoholsolution to get a 35% alcohol.

Solve the following problems using quadratic equations:

Example 4.5.4. Working together, Michael and Miguel can finish fixing a car intwo days. Working alone, Michael took three days longer than Miguel to finish thejob. How long did it take each to finish the job alone? Solution:

number of days part of job finished in 1 day

Michael x + 31

x + 3

Miguel x1

x

together 21

2

Since the part of the job that Michael and Miguel can together is the same as thecombined part of the job that they can finish individually, we have the followingequation:

1

x + 3+

1

x=

1

2

43

(Continuation)Solving this equation, we have

(2)(x)(x + 3)

(1

x + 3+

1

x

)= (2)(x)(x + 3)

(1

2

)(2)(x) + (2)(x + 3) = (x)(x + 3)

2x + 2x + 6 = x2 + 3x

4x + 6 = x2 + 3x

x2 x 6 = 0(x 3)(x + 2) = 0

Therefore x 3 = 0 or x + 2 = 0. Thus either x = 3 or x = 2 but since x stands forthe the number of days, x cannot be negative so we have x = 3 days. Thus, Miquelcan finish the job in 3 days while Michael can finish the job in x+3 = 3+3 = 6 days.

Example 4.5.5. The height of a triangle is 2cm more than its base. If its area is40cm2, what is the height of the triangle?Solution:Let x be the length of the base of the triangle. Thus the height of the triangle is

x + 2. Now recall that the area of a triangle, A =1

2bh. Hence we have the equation

40 =

(1

2

)(x)(x + 2) 40 =

(1

2

)(x2 + 2x) 40(2) = (x2 + 2x) 0 =

x2 + 2x 80. Observe that the resulting equation is quadratic and is factorable, thuswe can solve it by using Factoring Method. Note that factoring the right-hand sidegives us 0 = (x + 10)(x 8). Thus solving for x we get x = 10 and x = 8. Sincex represents the base and base is a nonnegative number, we have the base x = 8cm.Thus the height of the triangle is x + 2 = 8 + 2 = 10cm.

4.6 Inequalities

Definition 4.6.1. An inequality is a statement that one quantity or expression is greaterthan or less than another quantity or expression.The set of values that makes an inequality true is called a solution set.

To solve an inequality is to look for the values of the variable that satisfies the given inequality.Properties of InequalityFor any real numbers a, b, and c,

1. a b > 0 if a > ba b < 0 if a < b.

44

2. Closure Property of InequalityIf a > 0 and b > 0, then a + b > 0 and ab > 0.

3. Transitive Property of InequalityIf a < b and b < c, then a < c.

4. Continued InequalityIf a < b and b < c, then a < b < c.

5. Addition Property of InequalityIf a < b and c is a real number, then a + c < b + c.

6. Multiplication Property of InequalitySuppose a < b.Case 1: If c > 0, then ac < bc.Case 2: If c < 0, then ac > bc.

Remark 4.6.2. Instead of using sets to display the solution to an inequality an interval isoften used. The representations are as follow:

set notation interval notation{x < b} (, b){x b} (, b]{x > a} (a,){x a} [a,)

{a < x < b} (a, b){a x < b} [a, b){a < x b} (a, b]{a x b} [a, b]

Note that the symbol means infinity and that we use parenthesis (, ) when theinequality is strict, that is, it is either a < or a > relationship and a square bracket [, ]is used if the inequality is either or .

4.7 Linear Inequality in One Variable

To solve a linear inequality in one variable, apply the Addition Property of Inequality andthe Multiplication Property of Inequality.

45

Example 4.7.1. Solve the following linear inequalities.

1. 2x 4 > 3x + 3

2x 4 > 3x + 32x 3x > 3 + 4

x > 7x < 7

2.1

6x 3

4 3

8x +

1

2

24

(1

6x 3

4

) 24

(3

8x +

1

2

)4x 18 9x + 124x 9x 12 + 18

5x 30x 6

4.8 Quadratic Inequalities in One Variable

Quadratic Inequalities in One Variable are solved by using the Split-point Method.Steps:

1. Put all the terms of the inequalities on one side (preferably to the left) and write theleft-hand side expression in the form ax2 + bx + c.

2. By factoring, factor the left-hand side of the inequality and determine the roots of theleft-hand side of the inequality as if it is an equation.

3. Determine the intervals before and after each of the obtained solutions and analyze thevalues of the inequality in each interval. The interval/s which satisfy the inequalitywill be its solution set.

Example 4.8.1. Find the solution set of 3x2 16 < 2x.Solution:

3x2 2x 16 < 0

(3x 8)(x + 2) < 0

Roots:8

3and 2

46

(Continuation)Analysis:

Intervals (3x 8) (x + 2) (3x 8)(x + 2)x < 2 +

2 < x < 83

+

x >8

3+ +

Since the left-hand side of the inequality must be less than 0, the solution set is

2 < x < 83.

Example 4.8.2. Find the solution set of 2x2 7x 4 0.Solution:

2x2 7x 4 0

(2x + 1)(x 4) 0

Roots: 12

and 4

Analysis:Intervals (2x + 1) (x 4) (2x + 1)(x 4)

x 12

+

12 x 4 +

x 4 + + +Since the left-hand side of the inequality must be greater than 0, the solution set is

x 12

x 4.

47

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) The solution of the linear equation 4x 2 = 2 is

(a) x = 0 (b) x = 1 (c) x = 1 (d) x = 4

(2) The equation 4x + 5 = 3x 7 is true when x is equal to

(a)12

7(b) 2 (c) 12 (d) 2

7

(3) The value of x in the equation 3x 5 = 5 2x is

(a) 0 (b) 1 (c) 2 (d) 3

(4) The solution of the equation4

x + 2 2

x 2=

x 10x2 4

(a) 2 (b) 2 (c) 1 (d)

(5) The value of x in2

6x 7 5

3x 4= 0 is

(a)8

9(b)

1

8(c)

31

12(d)

9

8

(6) A plastic folder costs P25. How much will x folders cost if the total cost is representedby y?

(a) y = x + 25 (b) x = 25 + y (c) x + y = 25 (d) y = 25x

(7) Angie is five years older than Greg. Three years ago, she was twice as old as Greg.How old is Greg now?

(a) 6 (b) 8 (c) 13 (d) 15

48

(8) The solution set of |7x 5| = 9 is

(a)

{2,47

}(b)

{2, 4

7

} (c){

2,4

7

}(d)

{2, 4

7

}(9) Which of the following is a solution of the equation |x 1| = 2x?

(a) 1 (b) 13

(c)1

2(d) both a and b

(10) The solutions of the quadratic equation x(x 2) = 3 are

(a)

{3,3

2

}(b) {3,1} (c)

{3, 3

2

}(d) {3, 1}

(11) The roots of the equation x2 + 3x 10 = 0 are

(a) 5 and 2 (b) 10 and 1 (c) 5 and 2 (d) 5 and 2

(12) The sum of the roots of the quadratic equation 2x2 4x 5 = 0 is

(a) 52

(b)5

4(c) 2 (d) 2

(13) The product of the roots of the equation x2 3x + 2 = 0 is

(a) 2 (b) 2 (c) 3 (d) 3

(14) Which of the following is TRUE about the nature of the roots of 2x2 x 3 = 0?

(a) The roots are real and unequal.

(b) The roots are real and equal.

(c) The roots are imaginary and unequal.

(d) The roots are imaginary and equal.

(15) What is the value of the discriminant of the quadratic equation 5x2 3x + 2 = 0?

(a) D = 7 (b) D = 49 (c) D = 31 (d) D = 31

49

(16) Which quadratic equation has 1 +

3 and 1

3 for its roots?

(a) x2 + 2x 2 = 0(b) x2 2x + 2 = 0

(c) x2 2x 2 = 0(d) x2 + 2x + 2 = 0

(17) Using the method of completing the square in solving the equation x2 7x+4 = 0 theconstant term to be introduced is

(a)49

4(b)

9

4(c) 4 (d) none of these

(18) The solution of the linear inequality 4x 8 6x + 10 is

(a) x 2 (b) x 9 (c) x 9 (d) x 2

(19) The solution set of 3x + 1 > 6x 2 is

(a) x > 1 (b) x < 1 (c) x > 1 (d) x < 1

(20) The solution of the inequality 7 < 4 3x < 13 is

(a) 1 < x < 3 (b) 1 > x > 3 (c) 1 < x < 3 (d) 1 > x > 3

(21) If x + 1 < 11, then the solution set is

(a) (, 10) (b) (10,) (c) (10,) (d) (, 10)

(22) The solution set of the inequality 2x 3 5 is given by

(a) (, 4)(b) (4, +)

(c) (, 4](d) [4, +)

(23) The solution set of the inequality 1 3x5

< 0 is

(a)

(, 5

3

)(b)

(,5

3

) (c)(5

3,)

(d)

(5

3,)

50

(24) The solution set of 6x2 + 7x 20 0 is

(a)

(,5

2

][43, +

)(b)

(,4

3

][52, +

) (c)[5

2,4

3

](d)

[4

3,5

2

](25) The solution set to the inequality (x 1)2 0 is given by

(a) the set of real numbers

(b) (1,)(c) {1}(d)

51

CHAPTER 5

Systems of Linear Equations

5.1 Basic Concepts

Definition 5.1.1. A system of linear equations is a set of two or more linear equationsthat are considered simultaneously.

Three Kinds of Systems of Linear Equations

1. Consistent System - a system having a finite number of solutions.

2. Inconsistent System - a system that has no solution.

3. Dependent System - a system having an infinite number of solutions.

To solve a system of linear equations means to find its solution set.

Methods of Solving a System of Linear Equations

1. Substitution MethodSubstitution method is executed as follows:

(a) Solve one of the unknown in one of the equation in terms of the other unknowns.

(b) Substitute this expression in the other equations.

(c) Solve the resulting equation in step (b).

(d) Substitute the value obtained in step (c) back into the equation formed in step(a) to find the value of the other unknown.

2. Elimination MethodElimination method is done as follows:

(a) Multiply each equation in the system by an appropriate nonzero real number sothat one of the unknown can be eliminated by the process of addition.Note: This step may be unnecessary.

(b) Add the resulting equations from step (a) and solve for one of the unknown.

(c) Substitute the result in step (b) back in any of the original equations to find thevalue of the other unknowns.

52

5.2 Systems of Two Linear Equations in Two Unknowns

Definition 5.2.1. A system of two linear equations in two unknowns is a set of twolinear equations of the form ax + by + c = 0, where a, b, c are constants with a and b notsimultaneously zero unless c is also zero. In symbols, we write the system as

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

A solution to this system is an ordered pair that satisfies both equations.

Remark 5.2.2. Even without solving, we can identify the type of system the given is byusing the fact that:

(i) Ifa1a26= b1

b2;a1a26= c1

c2and

b1b26= c1

c2then the system is consistent or independent.

(ii) Ifa1a2

=b1b2

;a1a26= c1

c2and

b1b26= c1

c2then the system is inconsistent.

(iii) Ifa1a2

=b1b2

=c1c2

then the system is dependent.

Example 5.2.3. Solve

{x + y = 30x y = 20 .

Solution:

Solve this system of equations by substi-tution.First, we solve for y in (2). Doing so, weget y = x 20.Substitute y = x 20 to equation (1).

x + (x 20) = 302x 20 = 30

2x = 30 + 20

2x = 50

x = 25

To get the value of y corresponding tox = 25, we substitute x = 25 to y = x20and we get y = 2520 = 5. Hence, y = 5.Thus, we find that the solution to this sys-tem of equations is {(25, 5)}.

53

Example 5.2.4. Solve

{4x + 5y 2 = 02x 7y 20 = 0 .

Solution:

Multiply equation (2) by (-2).4x + 5y 2 = 0

(2) (2x 7y 20 = 0)

By the method of elimination, eliminatex by addition.

4x + 5y 2 = 0+ 4x + 14y + 40 = 0

19y + 38 = 0

Solve for y in the resulting equation.

19y + 38 = 0 19y = 38 y = 2.

Solve for x by substituting the obtainedvalue of y in equation (1).

4x + 5y 2 = 0 4x + 5(2) 2 = 0 4x 10 2 = 0 4x 12 = 0 4x = 12 x = 3.

Therefore, the solution to this system oflinear equations is {(3,2)}

5.3 Systems of Three Linear Equations in Three Unknowns

Definition 5.3.1. A system of three linear equations in three unknowns is a set ofthree linear equations of the form ax + by + cz + d = 0, where a, b, c, d are constants witha, b and c not all zero unless d is also zero. In symbols, we write the system as

a1x + b1y + c1z + d1 = 0

a2x + b2y + c2z + d2 = 0

a3x + b3y + c3z + d3 = 0

A solution to this system is an ordered triple that satisfies all equation.

54

Example 5.3.2. Solve

3x 5y + z 9 = 04x y + 2z 6 = 0x 2y + 3z 2 = 0

.

Solution:

3x 5y + z 9 = 0 (1)4x y + 2z 6 = 0 (2)x 2y + 3z 2 = 0 (3)

Using equations (1) and (2) eliminate zby multiplying (1) by (-2) then addingthe resulting equation to equation (2) toget equation (4).(2) (3x 5y + z 9 = 0 )

4x y + 2z 6 = 06x + 10y 2z + 18 = 0

4x y + 2z 6 = 02x + 9y + 12 = 0 (4)

Again, using equations (1) and (3) elimi-nate z by multiplying (1) by (-3) and thenadding the resulting equation to equation(3) to get equation (5).(3) (3x 5y + z 9 = 0 )

x 2y + 3z 2 = 09x + 15y 3z + 27 = 0

x 2y + 3z 2 = 08x + 13y + 25 = 0 (5)

Now using equations (4) and (5) eliminatex by multiplying (4) by (-4), then addingthe resulting equation to equation (5) toobtain a linear equation in y.(4) (2x + 9y + 12 = 0)

8x + 13y + 25 = 08x 36y 48 = 0

8x + 13y + 25 = 023y 23 = 0

Solve for y in the resulting equation.

23y 23 = 0 23y = 23 y = 1.

Solve for x by substituting the obtainedvalue of y in equation (4).

2x + 9y + 12 = 0 2x + 9(1) + 12 = 0 2x 9 + 12 = 0 2x = 9 12 2x = 3

x = 32

Solve for z by substituting the obtainedvalues of x and y in equation (3).

x 2y + 3z 2 = 0

(

3

2

) 2(1) + 3z 2 = 0

(

3

2

)+ 2 + 3z 2 = 0

3z = 32

z = 12

Therefore, the solution to this system of

linear equations is

{(3

2,1,1

2

)}.

55

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) Which of the following equations form a dependent system withx

2+

y

3= 1?

(a) 2x + 3y = 1 (b) 3x + 2y = 6 (c) x + y = 1 (d)x

2+

3

y= 6

(2) Given the system of equations

{x + y = 3x = y + 1

, which method is best applicable in finding

its solution?

(a) substitution (b) elimination (c) comparison (d) none of these

(3) What is the value of x in the solution to the system

{2x + y = 2x y = 4 ?

(a) 0 (b) 1 (c) 2 (d) 3

(4) What is the value of y in the solution to the system

{2x + y = 2x y = 4 ?

(a) 2 (b) 2 (c) 12

(d) 12

(5) What is the solution set of the system

{5x + 3y = 264x 9y = 2 ?

(a) {(2,4)} (b) {(4,2)} (c) {(2, 4)} (d) {(4, 2)}

(6) Which system of linear equations is described by 2x 3y = 8 and 6x 4y + 10 = 0?

(a) consistent

(b) dependent

(c) inconsistent

(d) cannot be determined

For numbers (7)-(10), consider the system of equations2x z = 1z 3y = 7x + y = 7

56

(7) Which method is best to use in the given system of equations given above?

(a) elimination

(b) substitution

(c) comparison

(d) cannot be determined

(8) What is the value of z in the solution to the given system?

(a) 3 (b) 4 (c) 5 (d)1

5

(9) What is the value of x in the system given above?

(a) 3 (b) 4 (c) 5 (d)1

5

(10) What is the value of y in the given system?

(a) 3 (b) 4 (c) 5 (d)1

5

57

CHAPTER 6

Relations and Functions

6.1 Domain and Range of a Relation

Definition 6.1.1. A relation is an association between two sets of objects which are pairedaccording to a pattern or a rule.A relation can be described in several ways, namely: arrowdiagrams, ordered pairs, tables, graphs, and mathematical sentences or equations.

Example 6.1.2.

(1) The sets of ordered pairs {. . . , (n,2n), . . . , (1, 2), (2, 4), (3, 6), (4, 8, , (n, 2n), }, {(1, 2), (1, 4), (1, 6), (1, 8)} and{(1,5), (2,5), (3,5), (4,5), (5,5)} are relations.

(2) Similarly, the sets {(x, y) : y = 3x}, {(x, y) : x2 y2 = 16} and {(x, y) : y =|x| 4} are relations.

Notice that in the first three examples the relations are described using orderedpairs, while the last three examples are described using mathematical sentences orequations.

Observe that the set {(1,5), (2,5), (3,5), (4,5), (5,5)} can be represented ina table as follow:

x 1 2 3 4 5y -5 -5 -5 -5 -5

.

Definition 6.1.3. The set containing all the first components of the ordered pairs describ-ing a relation is called the domain of the relation while the set containing all the secondcomponents is called the range of the relation.

Remark 6.1.4. When the relation is described by an equation or a rule to generate theordered pairs (x, y), the domain is obtained by solving for all the values of x that can beused to generate real values for y, while the range can be determined by determining all thepossible values of y that will result from using all the x values in the domain.

58

Example 6.1.5. Determine the domain and the range of each of the following:

1. {(1, 2), (2, 4), (3, 6), (4, 8)}

2. {. . . , (1, 2), (0, 2), (1, 2), (2, 2), . . .}

3. {(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)}

Solution:

1. Domain: {1, 2, 3, 4} Range: {2, 4, 6, 8}

2. Domain: The set of integers Range: {2}

3. Domain: {2} Range: {0, 1, 2, 3, 4}

Example 6.1.6. Determine the domain and the range of each of the relations definedby the following equations:

1. y = 3x

2. y = 2x2

3. y =

x 2

4. y =x + 2

x 1Solution:

1. Domain: The set of real numbers since there is no prohibited values for x.Range: The set of real numbers since there is no prohibited values for y.

2. Domain: The set of real numbers since there is no restriction to the possiblevalues for x.Range: The set of nonnegative even real numbers since 2x2 0 and is alwayseven.

3. Domain: Observe that the index of the radical is even, thus the radicand mustbe nonnegative. Hence, x 2 0, which implies that x 2. Therefore, theDomain= {x R : x 2}.For the range, note that since a square root is nonnegative, we find that y 0.Therefore the Range= {y R : y 0}.

59

4. Domain: Since the relation is defined by a rational expression, we find thatit is not defined for the values of x where the denominator is zero.Thus, thedenominator x1 6= 0, which implies that x 6= 1. Hence the Domain= {x R :x 6= 1}.Range: To determine the range, we solve for x in terms of y, we get

y =x + 2

x 1 y(x 1) = x + 2

yx y = x + 2 yx x = y + 2 x(y 1) = y + 2

x = y + 2y 1

Thus we find that y 6= 1 otherwise the denominator is also zero. Hence theRange= {y R : y 6= 1}.

6.2 Functions

Definition 6.2.1. A function is a special kind of relation. It is defined as a correspondencebetween two sets X and Y such that to each element x X there corresponds exactly oneelement y Y. It is also defined as a set of ordered pairs (x, y) such that no two distinctmembers have the same first element.

Example 6.2.2. Determine whether the given set is a function or a relation:

1. {(1, 1), (2, 2), (3, 3), (4, 4)}Answer: Function, since no two ordered pair have the same value for x butdifferent value for y.

2. {. . . , (1, 1), (0, 1), (1, 1), (2, 1), . . .} Answer: Function.

3. {(2, 0), (2, 1), (2, 2), (2, 3), (2, 4)}Answer: Relation since 2 has several values of y corresponding to it.

4. y = 5xAnswer: Function, since each value of x will yield exactly one value of y whensubstituted in the equation.

60

5. y = 2x2 Answer: Function.

6. x2 + y2 = 1Answer: Relation since for each value of x there are two values of y that wouldsatisfy the equation.

Remark 6.2.3. To evaluate a function for a specific value of x, replace each appearance ofx in the function with its given value and simplify the expression.

Example 6.2.4. Evaluate f(x) = 3x2 5x + 2 at x = 2.We simply replace x by 2 in the function and simplify the result.

f(2) = 3(2)2 5(2) + 2 = 3(4) 5(2) + 2 = 12 10 + 2 = 4.

Thus, f(2) = 4.

6.3 Operations on Functions

Given two functions f(x) and g(x).

1. The sum of f(x) and g(x) is defined by (f + g)(x) = f(x) + g(x).

2. The difference of f(x) and g(x) is defined by (f g)(x) = f(x) g(x).

3. The product of f(x) and g(x) is defined by (f g)(x) = f(x) g(x).

4. The quotient of f(x) and g(x) is defined by

(f

g

)(x) =

f(x)

g(x), where g(x) 6= 0.

5. The composition of f(x) and g(x) is defined by (f g)(x) = f(g(x))

Example 6.3.1. If f(x) = x + 1 and g(x) = x2 1, find

1. (f + g)(x) = f(x) + g(x) = (x + 1) + (x2 1) = x + 1 + x2 1 = x2 + x.

2. (f g)(x) = f(x) g(x) = (x + 1) (x2 1) = x + 1 x2 + 1 = 2 + x + x2.

3. (f g)(x) = f(x)g(x) = (x+1)(x21) = (x)(x2)+(x)(1)+(1)(x2)+(1)(1) =x3 + x2 x 1.

4.

(f

g

)(x) =

f(x)

g(x)=

(x + 1)

(x2 1)=

(x + 1)

(x + 1)(x 1)=

1

x 1.

5. (f g)(x) = f(g(x)) = f((x2 1)) = (x2 1) + 1 = x2.

61

6.4 Inverse Functions

Definition 6.4.1. A function f is said to be one-to-one if and only if whenever x1 and x2are in the domain of f,

f(x1) = f(x2) = x1 = x2or equivalently,

x1 6= x2 = f(x1) 6= f(x2).The first implication means that for a function f each x corresponds to exactly one y,

while the second implication means that different values of x correspond to different valuesof y.

Algebraic method of determining whether a function f is one-to-one

(1) Assume f(x1) = f(x2).

(2) Apply the defining equation on both members of the equation.

(3) Simplify the resulting equation.

(4) If the result gives x1 = x2 conclude that f is one-to-one, otherwise, conclude that f isNOT one-to-one.

Example 6.4.2. In each of the following, determine whether the given function isone-to-one.

1. f(x) = 3x + 1

f(x1) = f(x2)

3x1 + 1 = 3x2 + 1

3x1 + 1 + (1) = 3x2 + 1 + (1)3x13

=3x23

x1 = x2

Since x1 = x2, we conclude that f(x) = 3x + 1 is a one-to-one function.

2. G(x) = x3

G(x1) = G(x2)

x13 = x2

3

3

x13 =3

x23

x1 = x2

Since x1 = x2, we conclude that G(x) = x3 is a one-to-one function.

62

3. g(x) = x2 + 5

g(x1) = g(x2)

x12 + 5 = x2

2 + 5

x12 + 5 + (5) = x22 + 5 + (5)

x12 = x2

2x12 =

x22

|x1| = |x2|

But |x1| = |x2| does not necessarily imply that x1 = x2 since |x1| = |x2| x1 = x2, we conclude that g(x) = x2 + 5 is NOT a one-to-one function.

Definition 6.4.3. If f is a one-to-one function then there is a function f1 defined as

x = f1(y) if and only if y = f(x).

If f is the set of ordered pairs (x, y) then f1 is the set of ordered pairs (y, x).

Remark 6.4.4. From the definition, the domain of f1 is the range of f, while the rangeof f1 is the domain of f.

Moreover, for any function f and its inverse f1,

(f f1)(x) = x for every x in the domain of f1.

(f1 f)(x) = x for every x in the domain of f.

Note that only one-to-one functions have inverses. Thus otherwise stated you have tocheck first whether the function is one-to-one before finding its inverse.

Procedure For Finding f1

Given a function y = f(x).

(1) Interchange x and y in the equation y = f(x). Thus the equation becomes x = f(y).

(2) Solve for y in the resulting equation.

(3) Set y = f1(x).

63

Example 6.4.5. Determine the inverse of the following one-to-one functions.

1. f(x) = 3x + 1

y = f(x)

y = 3x + 1

x = 3y + 1

x 1 = 3yx 1

3=

3y

3

y =x 1

3

f1(x) =x 1

3

To check whether the inverse function obtained is correct, show that (f f1)(x) = x and (f1 f)(x) = x. Observe that (f f1)(x) = f(f1(x)) =

f

(x 1

3

)= 3

(x 1

3

)+ 1 = (x 1) + 1 = x. Moreover, (f1 f)(x) =

f1(f(x)) = f1 (3x + 1) =(3x + 1) 1

3=

3x

3= x. Therefore, the inverse

function of f(x) = 3x + 1 is indeed f1(x) =x 1

3.

2. g(x) = 3

x + 1

y = 3

x + 1

x = 3

y + 1

(x)3 =(

3

y + 1)3

x3 = y + 1

x3 1 = yg1(x) = x3 1

Observe that g1(x) = x31 is the inverse of g(x) = 3

x + 1 since (gg1)(x) =g(x3 1) = 3

(x3 1) + 1 = 3

x3 = x, and (g1 g)(x) = g1

(3

x + 1)

=(3

x + 1)3 1 = (x + 1) 1 = x.

64

Practice Exercises

Write the letter corresponding to your answer on your answer sheet.

(1) What do you call a set of ordered pairs where no two elements have the same firstcoordinate?

(a) domain (b) function (c) range (d) relation

(2) Which of the following statements about function is ALWAYS TRUE?

(a) The domain and range have equal number of elements.

(b) Some elements of the domain have corresponding values in the range.

(c) Every element of the domain has a corresponding value in the range.

(d) None of the above.

(3) The given table below is

x 1 1 0 3 5 6y 1 2 3 4 2 3

(a) not a function (b) a function (c) a domain (d) a range

(4) If the domain of y = 3x 2 is {2,1, 0, 1, 2}, what is its range?

(a) {1, 2, 4, 5, 8}(b) {1, 4,2,5,8}

(c) {1,4, 2, 5, 8}(d) {1,2,4,5,8}

(5) The domain of the function f(x) =4

x 4is the set of

(a) all real numbers

(b) all real numbers except 4(c) all real numbers except 4

(d) all real numbers except 0

(6) Which of the following equations defines a function?

(a) y2 + (x + 1)2 = 4

(b) x + y = 5

(c) y2 = x 3(d) x + y2 = 0

65

(7) The following are functions EXCEPT

(a) {(1, 2), (6, 7), (3, 5), (9, 0)}(b) {(8, 9), (2, 3), (5, 8), (8, 8)}

(c) {(9, 2), (3, 9), (1, 9), (0, 1)}(d) {(0, 1), (1, 2), (2, 3), (3, 4)}

(8) The range of y =x

2x 1is

(a) all real numbers except 0

(b) all real numbers except1

2

(c) all real numbers except 0 and1

2(d) none of the above

(9) The domain of the function f(x) =

x 3 is

(a) {x : x > 3} (b) {x : x 3} (c) {x : x 3} (d) {x : x < 3}

(10) Which is NOT a function?

(a) y = 3x + 4 (b) x2 + y2 = 1 (c) y = x2 (d) y = 9

(11) What equation represents the relation shown by the table of values below?

x 1 2 3y 0 1 2

(a) x + y = 1 (b) x y = 1 (c) x + y = 1 (d) x y = 1

(12) What is the domain of the relation {(0, 0), (1, 1), (1,1), (2, 4), (2,4), (3, 9)}?

(a) {}(b) {4,1, 0, 1, 2, 4, 9}

(c) {0, 1, 2, 3}(d) {4,1, 0, 1, 4, 9}

(13) Which is NOT a one-to-one function?

(a) y = 5x 3 (b) y = x2 + 3x (c) y = 1x + 1

(d) y = 3x + 4

(14) If f(x) = 4x + 21 then f1(x) is equal to

(a) f1(x) =x 21

4

(b) f1(x) = x 214

(c) f1(x) =x

4 21

(d) f1(x) =1

4x + 21

(15) If f(x) = 3x + 1 then f [f1(7)] is equal to

66

(a) does not exist

(b)1

22

(c) 7

(d) none of these

For numbers (16)-(19), consider the following functions: f(x) = 3x + 1; g(x) = 2x 5; h(x) = 6x2 13x 5.

(16) (f + g)(x) is equal to

(a) 5x + 4 (b) 5x + 3 (c) 5x 4 (d) 5x 4

(17) (g f)(x) is equal to

(a) x + 6 (b) x + 6 (c) x 6 (d) x 6

(18)

(h

g

)(x) is equal to

(a) 3x + 1 (b) 2x + 3 (c) 3x 1 (d) 2x 3

(19) (h g)(x) is equal to

(a) 14x + 30(b) 24x2 146x + 210

(c) 14x 30(d) 24x2 146x + 80

(20) Given f(x) = x + 2 and g(x) = x2 4, find (g f)(1).

(a) 5 (b) 5 (c) 1 (d) 1

67

Comprehensive Examination

Write the letter corresponding to your answer.

(1) Which of the following pairs of algebraic expressions are similar?

(a) 4x y and 4xy

(b) 3rs and 2rs

(c)a

band ab

(d) 7m2n and 3mn2

(2) Which of the following expressions is a polynomial?

(a)5

x+ 3 (b)

x

5+ 3 (c) x

12 + 1 (d)

x + 2

(3) Which of the following expressions is NOT a binomial?

(a) 4x(x + 1) (b)x + 1

2(c) 4x + 5 (d) (2x 1)2

(4) What is the degree of the term 3x2yz?

(a) 2 (b) 4 (c) 5 (d) 3

(5) What is the degree of x4y3z2 + x6y + xz yz5?

(a) 6 (b) 2 (c) 7 (d) 9

(6) The sum of 3x y + 2z, 3y 6z + 7x and 9z 8x 2y is

(a) 2x + 5z

(b) 18x + 7y 17z(c) 16x y + 13z(d) 2x 5z

(7) The result of subtracting 6a b from 4a + b is

(a) 2b 2a(b) 2a 2b

(c) 2a

(d) 10a

68

(8) Which of the following is the sum when the difference between 20m2 + 4t xh2 and2m2 + 10xh2 8t is added to 6xh2 + 13t 8m2?

(a) 10m2 + 5xh2 + t

(b) 10m2 + 9t + 5xh2

(c) 10m2 + 25t 17xh2

(d) 14m2 + 25t 17xh2

(9) What is the product of (3x 4) and (2x 5)?

(a) 6x2 7x + 20 (b) 6x2 + 7x + 20 (c) 6x2 23x + 20 (d) 6x2 23x 20

(10) (4x5)3 is equal to

(a) 4x15 (b) 4x8 (c) 64x15 (d) 64x15

(11) The reciprocal of (3x + 5) is

(a) (3x 5) (b) (5 + 3x) (c) 1(3x + 5)

(d) 1(3x + 5)

(12) The quotient when 18x2y3 is divided by 6xy is

(a) 3xy2 (b) 12x3y4 (c) 24x3y4 (d) 6x2y

(13) One of the factors of a2 7a 18 is a + 2. What is the other factor?

(a) (a 9)(b) (a 6)

(c) (a 4)(d) (a 2)

(14) The product of (x3 + 3y2)(x3 3y2) is

(a) x9 9y4 (b) x6 9y4 (c) x9 + 9y4 (d) x6 + 9y4

(15) Which is the product of (3x + y)3?

(a) 27x3 + y3

(b) 27x3 9x2y + y3(c) 27x3 + 27x2y + 9xy2 + y3

(d) 27x3 27x2y + 9xy2 y3

(16) The square of (2ab cd) is equal to

(a) 4a2b2 c2d2

(b) 2abcd

(c) 4a2b2 4abcd + c2d2

(d) none of these

69

(17) (2a3 + 3b2 5c)2 is equal to

(a) 4a6 + 9b4 + 25c2 + 6a3b2 10a3c 15b2c(b) 4a6 + 9b4 + 25c2 + 12a3b2 20a3c 30b2c(c) 4a9 + 9b4 + 25c2 + 12a3b2 20a3c 30b2c(d) 4a9 + 9b4 + 25c2 + 6a3b2 10a3c 15b2c

(18) Which of the following is not a rationally factorable quadratic trinomial?

(a) 2a2 + 5a + 9 (b) 10a2 + 13a + 4 (c) 4a2 + 11a + 6 (d) 3a2 + 7ab + 4b2

(19) 4p2 16pq + 16q2 can be factored as

(a) (2p + 4q)2

(b) (2p 4q)2(c) (2p 4q)(2p + 4q)(d) (4p 2q)(4p + 2q)

(20) The factors of the polynomial x2 125

are

(a)

(x 1

5

)and

(x +

1

5

)(b) (x 1) and

(x 1

25

) (c) (x 5) and (x 5)(d) (x + 5) and (x + 5)(21) What is the complete factorization of x4 y4?

(a) (x2 + y2)(x y)(x + y)(b) (x2 y2)(x2 + y2)

(c) (2x + 2y)(x + y)(x y)(d) (2x + 2y)(x2 + y2)

(22) The factors of 8x2 + 2x 15 are

(a) (2x + 3) and (4x 5)(b) (8x + 3) and (x 5)

(c) (2x + 15) and (4x 1)(d) (2x 3) and (4x + 5)

(23) x 2 is a factor of the following EXCEPT

(a) x2 + x 6(b) x2 + 5x + 6

(c) x2 5x + 6(d) x2 4

70

(24) The complete factorization of 3m3 9m2 + 6m is

(a) (m2 2m)(m 3)(b) (3m)(m 2)(m 1)

(c) (3m2 2)(m 3)(d) (3m)(m2 3m + 2)

(25) The factors of (2x y)3 8 are

(a) [(2xy)2][(2x y)2+2(2xy)+4](b) [(2xy)2][(2x y)22(2xy)+4]

(c) [(2xy)2][(2x y)24(2xy)+4](d) [(2xy)2][(2x y)2+4(2xy)+4]

(26) An expression where a fraction is within a fraction is called

(a) rational expression

(b) algebraic expression

(c) complex fraction

(d) complex numbers

(27) The fractionx3 y3

x2 y2is equivalent to

(a) x + y (b)x2 + xy + y2

x + y(c) x y (d) x

2 xy + y2

x + y

(28) Which of the following fractions cannot be reduced to lowest term?

(a)(a + 2)(a 4)(a + 2)(a 4)

(b)(x + 5) + y

(x + 5) + c

(c)x + 5 + y

x + 5 + y

(d)3(x + 4)

6(x + 4)

(29) Which is equivalent to1

a 1?

(a)a2

a2 1

(b)a 1a2 1

(c)a2 + a + 1

a3 1

(d)a2 a + 1

a3 1

(30)x + 4

16x + 4x2when reduced to lowest term is

(a) 14x

(b) 4 + x

(c)1

4x

(d) 4 + x2

(31) The simplified form of14m4n3z

7m2nis

71

(a) 2m2n2z (b) 2mnz (c) 2mnz (d) 2m2n2z

(32) The LCD if the denominators are (x 2)4(x + 1)3(x 1)2, (x 2)(x + 1)3(x 1) and(x 2)2(x 1)2

(a) (x 2)4(x + 1)3(x 1)2

(b) (x 2)(x + 1)(x 1)(c) (x 2)7(x + 1)4(x 1)3

(d) (x 1)

(33) The sum of3x + 1

x2 1+

2

x + 1+

3

x 1is

(a)10x

x2 1(b)

2x + 8

x2 1(c)

8x + 2

x2 1(d)

6x

x2 1

(34) The difference when2

8 xis subtracted from

3

x 8is

(a)5

x 8

(b)1

8 x

(c)1

x + 8

(d)5

8 x

(35)3

a+

2

a2+

1

a3is equal to

(a)6

a3

(b)6

a + a2 + a3

(c)a3 + 2a2 + 3a

a3

(d)1 + 2a + 3a2

a3

72

(36) The simplified form ofr + s

rs 1

s+

1

r sis

(a)2

r(b)

2r sr s

(c)2r s

r(r s)(d) 2

(37)n

d r

sis equal to

(a)ns

dr (b)ds

nr(c)

nr

ds (d)dr

ns

(38)

[x yx + y

][x2 + xy

xy2

]

[(x y)3

y2

]is equal to

(a)1

(x y)2

(b)y4

(x y)4