Question

using the factor theorem show that g(x) is a factor of the p(x) when p(x)= 2 x to the power 4 + 9 x to the power 3 + 6x to the power 2 -11x -6, g(x)= x-1​

Answer 1

Given,

p(x) = 2x⁴ + 9x³ + 6x² - 11x - 6

And,

g(x) = x - 1

We have to show that g(x) is a factor of p(x).

On considering g(x), we can say that g(x) = 0 for x = 1, i.e., it's nothing but g(1) = 0.

Then, let's check whether p(1) = 0. It will imply that g(x) can exactly divide p(x) if it is. So,

p(1) = 2(1)⁴ + 9(1)³ + 6(1)² - 11(1) - 6

p(1) = 2 + 9 + 6 - 11 - 6

p(1) = 0

Hence it's true that g(x) is a factor of p(x).

Well,

p(x) = 2x⁴ + 9x³ + 6x² - 11x - 6

p(x) = 2x⁴ - 2x³ + 11x³ - 11x² + 17x² - 17x + 6x - 6

p(x) = 2x³(x - 1) + 11x²(x - 1) + 17x(x - 1) + 6(x - 1)

p(x) = (x - 1)(2x³ + 11x² + 17x + 6)

p(x) = g(x) · (2x³ + 11x² + 17x + 6)

This implies g(x) can exactly divide p(x).

#answerwithquality

#BAL