Question

What is the remainder R when the polynomial p(x) is divided by (x - 2)? Is (x - 2) a factor of p(x)? P(x) = -4x4 + 6x3 + 8x2 - 6x - 4 A) R = 0, no B) R = 0, yes C) R = -72, no D) R = -72, yes

Answer 1

Answer:

B

(x-2) is a factor since remainder is 0.

Step-by-step explanation:

We divide (x-2) into the polynomial $-4x^4+6x^3+8x^2-6x-4$ through long division or synthetic. We choose long division and look for what will multiply with (x-2) to make the polynomial $-4x^4+6x^3+8x^2-6x-4$ .

$(x-2)(-4x^3)=-4x^4+8x^3$

We subtract this from the original $-4x^4-(-4x^4)+6x^3-(8x^3)+8x^2-6x-4$.

This leaves $-2x^3+8x^2-6x-4$. We repeat the step above.

$(x-2)(-2x^2)=-2x^3+4x^2$.

We subtract this from $-2x^3-(-2x^3)+8x^2-(4x^2)-6x-4=4x^2-6x-4$. We repeat the step above.

$(x-2)(4x)=4x^2-8x$.

We subtract this from $4x^2-(4x^2)-6x-(-8x)-4=2x-4$. We repeat the step above.

$(x-2)(2)=2x-4$.

We subtract this from $2x-(2x)-4-(4)=0$. There is no remainder. This means (x-2) is a factor.