1001 Algebra Problems.PDF

Set 28 (Page 74)

433. c.First, factor the polynomial:

x^2 – 36 = x^2 – 6^2 = (x– 6)(x+ 6)

Now, set each of the factors equal to zero and

solve for xto conclude that the zeros of the

polynomial are –6 and 6.

434. a.First, factor the polynomial:

9 x^2 – 25 = (3x)^2 – (5)^2 = (3x– 5)(3x+ 5)

The factors are 3x– 5 and 3x+ 5. Now, set

each factor equal to zero and solve for x.The

zeros of the polynomial are –^53 and ^53 .

435. d.First, note that 5x^2 + 49 cannot be factored
     further. Since both terms are positive, the sum
     is positive, so there is no x-value that makes
     the expression equal to zero.


436. b.First, factor the polynomial:

6 x^2 – 24 = 6(x^2 ) – 6(4) = 6(x^2 – 4) = 6((x)^2 ) =

6(x– 2)(x+ 2)

The factors are 6,x – 2, and x+ 2. Set each fac-

tor equal to zero and solve for xto conclude

that the zeros of the polynomial are –2 and 2.

437. d.Begin by factoring, the polynomial:

5 x(2x + 3) – 7(2x+ 3) = (2x+ 3)(5x– 7)

The factors are 2x + 3 and 5x– 7. Now, set

each factor equal to zero and solve for xto find

that the zeros of the polynomial are –^32 and ^75 .

438. d.First, factor the polynomial:

5 x(^23 x+ 7) – (^23 x+ 7) = (5x–1)(^23 x+ 7)

The factors are 5x– 1 and ^23 x+ 7. Now, set

each factor equal to zero and solve for x.The

zeros of the polynomial are ^15 and –^221 .

439. a.Begin by factoring the polynomial:

28 x(5 – x) – 7x^3 (5 – x) = (28x– 7x^3 )(5 – x) =

7 x(4 – x2)(5 – x)

= 7x(2^2 – x^2 )(5 – x) = 7x(2 – x)(2 + x)(5 – x)

There are four factors: 7x, 2 – x, 2 + x, and 5 –

x. Now, set each of these factors equal to zero

and solve for x.The zeros of the polynomial

are 0, –2, 2, and 5.

440. c.First, factor the polynomial:

75 x^4 + 30x^3 + 3x^2 = 3x^2 [25x^2 + 10x+ 1] =

3 x^2 [25x^2 + 5x+ 5x+ 1]

= 3x^2 [5x(5x+ 1) + (5x+ 1)] = 3x^2 [(5x+ 1)

(5x+ 1)] = 3x^2 (5x+ 1)^2

The factors are 3x^2 and (5x+ 1)^2 .Now,set

each factor equal to zero and solve for xto

conclude that the zeros of the polynomial are

0 and –^15 .

441. a.First, factor the polynomial:
     x^2 – 9x+ 20 = x^2 – 5x– 4x+ 20 =
     (x^2 – 5x) – (4x– 20) = x(x– 5) = (x– 4)(x– 5)

Now, set each factor on the right side of the

string of equalities equal to zero and solve for

x.The zeros of the polynomial are 4 and 5.

442. c. Begin by factoring the polynomial:

12 x^2 – 37x– 10 = 12x^2 + 3x– 40x– 10 =

3 x(4x+ 1) – 10(4x+ 1) = (3x– 10)(4x+ 1)

The factors are 3x– 10 and 4x+ 1. Now, set

each factor equal to zero and solve for xto find

that the zeros of the polynomial are ^130 and –^14 .

ANSWERS & EXPLANATIONS–