1001 Algebra Problems.PDF

number in each subinterval, and record the

sign of the expression above each:

Since the inequality includes “equals,” we

include those values from the number line

that make the polynomial equal to zero.

The solution set is[–2, –1]∪[0,∞).

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

–4x^5 + 24x^4 – 20x^3 = 4x^3 (x^2 – 6x+ 5) =

–4x^3 (x^2 – x– 5x+ 5)

= –4x^3 (x(x– 1) – 5(x– 1)) = –4x^3 (x– 5)(x– 1)

Next, set each factor equal to zero and solve

for x.The zeros of the polynomial are 0, 1, and

5. Now, we assess the sign of the expression on
   the left side on each subinterval formed using
   these values: We form a number line, choose a
   real number in each subinterval, and record
   the sign of the expression above each:

The inequality includes “equals,” so we include

those values from the number line that make

the polynomial equal to zero. The solution set

is(–∞,0]∪[1, 5).

462. a.First, determine the x-values that make the
     expression on the left side equal to zero. This
     requires that we factor the polynomial:

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

x(x^2 – 4) – (x^2 – 4) = (x^2 – 4)[2x^2 – x–1]

= (x^2 – 4)[2x^2 – 2x+ x– 1] = (x^2 – 4)[2x(x– 1)

+ (x– 1)] = (x^2 – 4)(2x+ 1)(x– 1)

= (x– 2)(x+ 2)(2x+ 1)(x– 1)

Set each factor equal to zero and solve for xto

find the zeros of the polynomial are 1, 2, –2,

and –^12 . 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:

The inequality does not include “equals,” so we

exclude those values from the number line that

make the polynomial equal to zero. The solu-

tion set is (–2, –^12 )∪(1, 2).

463. c.Determine the x-values that make the
     expression on the left side equal to zero. First,
     factor the polynomial:

2 x^2 (16 + x^4 ) + 3x)16 + x^4 ) + (16 + x^4 ) =

(16 + x^4 )[2x^2 + 3x+ 1]

= (16 + x^4 )[2x^2 +2x+ x+ 1] = (16 + x^4 )[2x(x+

1) + (x+ 1)] =(16 + x^4 )(2x+ 1)(x+ 1)

Set each factor equal to zero and solve for x.

The zeros of the polynomial are – 1 and –^12 .

Assess the sign of the expression on the left

side on each subinterval formed using these

(^212)

• – + – +
  1

• 2

0 15

+ – +–

2 10

• 
  
  
  
  • –+
    ––
  
  
  
  

ANSWERS & EXPLANATIONS–