Question

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

A) p(x)=2x³+x²−2x−1, g(x)=x+1
B) p(x)=x³+3x²+3x+1, g(x)=x+2
C) p(x)=x³−4x²+x+6, g(x)=x−3

Answer 1

To determine whether g(x) is a factor of p(x) using the Factor Theorem, we need to check if p(x) is divisible by g(x).

Step-by-step explanation:

To determine whether g(x) is a factor of p(x) using the Factor Theorem, we need to check if p(x) is divisible by g(x). Here are the steps for each case:

A) p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1

Plug in g(x) = x + 1 into p(x), we get p(-1) = 2(-1)³ + (-1)² - 2(-1) - 1 = 0. Since p(-1) = 0, g(x) = x + 1 is a factor of p(x).

B) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

Plug in g(x) = x + 2 into p(x), we get p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1 = 1. Since p(-2) ≠ 0, g(x) = x + 2 is not a factor of p(x).

C) p(x) = x³ - 4x² + x + 6, g(x) = x - 3

Plug in g(x) = x - 3 into p(x), we get p(3) = 3³ - 4(3)² + 3 + 6 = 54. Since p(3) ≠ 0, g(x) = x - 3 is not a factor of p(x).