Question 2(5 points)

Findp(-5) andp(3) for the functionp(x) = 2x⁵- 9x⁴- 2x²+ 12x- 2.

Question 2 options:

-11,987; -227

-11,915; -251

-4,487; -551

-11,985; -225

Question 3(5 points)

Use synthetic substitution to findg(3) andg(-5) for the functiong(x) =x⁵- 8x³- 2x+ 7.

Question 3 options:

28, -2,108

40, -2,128

460, 2,122

-134, 1,642

Question 4(5 points)

Expand the given power using the Binomial Theorem.

(z- 11)⁴

Question 4 options:

z⁴- 44z³+ 726z²- 5324z+ 14641

z⁴+ 44z³+ 726z²+ 5324z+ 14641

z⁴- 44z³+ 726z²- 44z+ 1

11z⁴+ 44z³+ 726z²+ 5324z+ 14641z

Question 5(5 points)

Simplify the expression using long division.

(10x²- 85x- 10) (x- 8)

Question 5 options:

quotient 10x- 5 and remainder -50

quotient 10x- 85 and remainder 8

quotient 10x- 5 and remainder -30

quotient 10x+ 5 and remainder 30

Question 6(5 points)

Factor the polynomial completely.

12a⁴b²- 18a³b²

Question 6 options:

6(2a⁴b²- 3a³b²)

6a³b²(2a- 3)

a³b²(12a- 18)

6a²b²(2a²- 3)

Question 7(5 points)

Findp(2) andp(4) for the functionp(x) = 6x⁴+ 4x³- 3x²+ 8x+ 15.

Question 7 options:

147; 1,791

99; 639

139; 1,767

132; 1,776

Question 8(5 points)

For the given function, determine consecutive values of x between which each real zero is located.

f(x) = -11x⁴- 5x³- 9x²+ 12x+ 10

Question 8 options:

There are zeros between x=2andx=3,x=1andx=2,0 x=-1andx=-2,x=-1andx=-2,x=-2andx=-3.

There are zeros between 00 x=1andx=0,x=0andx=-100.

There is a zero between x=0andx=-100.

There is a zero between 00x=0andx=1.

Question 9(5 points)

Simplify the expression using synthetic division.

(6x³- 84x²+ 264x- 240) (x- 10)

Question 9 options:

quotient 6x²- 144x-1176 and remainder 11,520

quotient 66x²+ 576x- 6,024 and remainder 60,000

quotient 6x²- 24x+ 24 and remainder 0

quotient 60x²+516x+ 5,424 and remainder 54,000

Question 10(5 points)

Expand the given power using the Binomial Theorem.

(10k-m)⁵

Question 10 options:

100,000k⁵+ 50,000k⁴m+ 10,000k³m²+ 1,000k²m³+ 50km⁴+m⁵

100,000k⁵- 50,000k⁴m+ 10,000k³m²- 1,000k²m³+ 50km⁴-m⁵

100,000m⁵+ 50,000km⁴+ 10,000k²m³+ 1,000k³m²+ 50k⁴m+k⁵

k⁵- 5k⁴m+ 10k³m²- 10k²m³+ 5km⁴-m⁵

Question 12(5 points)

Simplify the given expression. Assume that no variable equals 0.

(17x^-5y¹¹)(-2xy⁸)

Question 12 options:

-34x^-4y¹⁹

-34y¹⁹

x⁴

-34x¹⁹y^-60

15y¹⁹

x⁴

Question 14(5 points)

Use synthetic substitution to findg(4) andg(-5) for the functiong(x) = 3x⁴- 5x²+ 7x- 7.

Question 14 options:

723, 1,778

133, -292

709, 1,708

869, 472

Question 15(5 points)

Simplify the given expression.

(16x²+ 15xy- 19y²) - (3x²- 3xy)

Question 15 options:

13x²- 18xy

13x²+ 12xy -19y²

13x²- 15xy- 16y²

13x²+ 18xy- 19y²

Question 16(5 points)

Simplify the given expression.

-10xy(4xy³- 7xy+ 9y²)

Question 16 options:

-40x²y⁴-7x²y²+ 9x²y³

-40x²y⁴+ 70xy+ 90y²

-40x²y⁴+70x²y²-90xy³

-40x²y⁴-7xy+ 9y²

Question 17(5 points)

Simplify the given expression.

5a³(9ab³- 5a²b²+ 2a³b)

Question 17 options:

45a⁴b³-25a⁵b²+ 10a⁶b

45a²b⁴-25a²b²+ 2a²b³

45a⁴b³-25a⁵b²+ 10a⁶b³

45a⁴b³-25a²b²+ 10a³b³

Question 18(5 points)

Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some of the factors may not be binomials.

16x³- 48x²- 49x+ 147;x- 3

Question 18 options:

(4x- 7)(4x+ 7)

(16x²- 49)

(4x- 7)

(4x- 7)(4x- 7)

Question 19(5 points)

Simplify the expression using long division.

(3x²- 97x+ 32) (x- 32)

Question 19 options:

quotient 3x- 97 and remainder 32

quotient 3x- 1 and remainder 0

quotient 3x- 1 and remainder -64

quotient 3x+ 1 and remainder 64