1001 Algebra Problems.PDF

values: Form a number line, choose a real

number in each of the subinterval, and record

the sign of the expression above each, as

follows:

The inequality includes “equals,” so we include

those values from the number line that make

the polynomial equal to zero. Therefore, the

solution set is [–1, –^12 ].

464. b.Find the x-values that make the expression
     on the left side equal to zero. First, factor the
     polynomial:

18(x^2 + 6x+ 8) – 2x^2 (x^2 + 6x+ 8) = (x^2 +

6 x+ 8)[18 – 2x^2 ] = (x^2 + 6x+ 8)[2(9 – x^2 )]

= (x^2 + 4x+ 2x+ 8)[2(3^2 – x2)] = (x(x + 4) +

2(x+ 4))[2(3 – x)(3 + x)]

= 2(x+ 2)(x+ 4)(3 – x)(3 + x)

Set each factor equal to zero and solve for xto

find the zeros of the polynomial, which are –4,

–2, –3, and 3. Assess the sign of the expression

on the left side on each subinterval formed using

these values. To this end, form a number line,

choose a real number in each subinterval, and

record the sign of the expression above each:

Because the inequality does not include

“equals,” we exclude those values from the

number line that make the polynomial equal

to zero. The solution set is(–4, –3)∪(–2, 3).

Section 4—Rational Expressions

Set 30 (Page 78)

465. d. = =


466. d. = = ^1 x


467. a. = = =


468. b.y

2

8

6

y

^4 = (y–8

8

)(

y

y

+8)== –(y+ 8)

469. a. = = =


470. c. = =

= =

471. a.A rational expression is undefined at any
     value ofxthat makes the denominator equal
     to zero even if the corresponding factor can-
     cels with one in the numerator. Observe that
     the denominator factors as 4x^3 + 44x^2 + 120x
     = 4x(x^2 + 11x+ 30) = 4x(x+ 5)(x+ 6).

Setting each factor equal to zero shows that

the rational expression is undefined at x= 0, –5,

and –6.

472. c.The domain of a rational expression is the
     set of all real numbers that do not make the
     denominator equal to zero. For this function,
     the values ofxthat must be excluded from
     the domain are the solutions of the equation
     x^3 – 4x= 0. Factoring the left side yields the
     equivalent equation

x^3 – 4x= x(x^2 – 4) = x(x– 2)(x+ 2) = 0

The solutions are x= –2, 0, and 2. Hence, the

expression is defined for any xin the set

(–∞,–2)∪(–2,0) ∪(0,–2) ∪(2,∞).

^1

2 x– 12

^1

2(x– 6)

^2 x(x+ 2)

4 x(x– 6)(x+ 2)

^2 x(x+ 2)

4 x(x^2 – 4x– 12)

^2 x^2 + 4x

4 x^3 – 16x^2 – 48x

^1

x– 8

x(x+ 8)

x(x+ 8)(x– 8)

x(x+ 8)

x(x^2 – 64)

x^2 + 8x

x^3 – 64x

(y– 8)(y+ 8)

–(y– 8)

z(z+ 4)

8

z(z–4)(z+4)

8(z–4)

z(z^2 – 16)

8(z– 4)

z^3 – 16z

8 z– 32

^25 x^4

x 25 x^4

25(–x)^4

x(5x^2 )^2

^2 z+ 5

z+ 5

(2z+ 5)(z– 3)

(z+ 5)(z– 3)

^2 z^2 – z– 15

z^2 + 2z– 15

4 3 2 3

––+

• ––

+ –

1 1

2

+ – +

• –

ANSWERS & EXPLANATIONS–