Question

One factor of f(x)=4x^3-4x^2-16x+16 is (x – 2). What are all the roots of the function? Use the Remainder Theorem.

A. x=1, x=2, or x=4
B. x=-2, x=1, or x=2
C. x=2, x=4, or x=16
D. x=-16, x=2, or x=16

Answer 1

It's B

Explanation:

Answer 2

One of the factor is (x-2).

we can apply the remainder theorem as below

`(4x^3-4x^2-16x+16)/((x-2))=(4x^2(x-1)-16(x-1))/((x-2))\\  \\(4x^2(x-1)-16(x-1))/((x-2))=((x-1)(4x^2-16))/((x-2))\\ \\((x-1)(4x^2-16))/((x-2))=(4(x-1)(x+2)(x-2))/((x-2))\\\\`

Hence the factors are (x+2),(x-1) and (x-2).

Hence the correct option is B.