Question

Questions 7 and 8 multiple choice

Answer 1

Problem 7

Answer: Choice A) 7

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Work Shown:

p(x) = x^4+x^3-kx-x+6

If (x-2) is a factor of p(x), then x = 2 is a root of p(x). This is a special case of the remainder theorem.

This means p(2) = 0

Replace every x with 2 and solve for k

p(x) = x^4+x^3-kx^2-x+6

p(2) = 2^4+2^3-k(2)^2-2+6

p(2) = 16+8-4k-2+6

p(2) = 28-4k

0 = 28-4k .... replace p(2) with 0

4k = 28

k = 28/4

k = 7

The polynomial

p(x) = x^4+x^3-7x^2-x+6

has the factor (x-2)

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Problem 8

Answer: D) 40

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Work Shown:

Check out the diagram below to see the synthetic division table. We're after the remainder which is highlighted in yellow.

An alternative is to use direct substitution

p(x) = 2x^4+4x^3-x^2-5

p(-3) = 2(-3)^4+4(-3)^3-(-3)^2-5 ... replace every x with -3

p(-3) = 2(81)+4(-27)-9-5

p(-3) = 162-108-9-5

p(-3) = 40  

This helps confirm we got the right remainder and the right answer.