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

Solution:

One of the factor is (x-2).

we can apply the remainder theorem as below

$\frac{4x^3-4x^2-16x+16}{(x-2)}=\frac{4x^2(x-1)-16(x-1)}{(x-2)}\\  \\\frac{4x^2(x-1)-16(x-1)}{(x-2)}=\frac{(x-1)(4x^2-16)}{(x-2)}\\ \\\frac{(x-1)(4x^2-16)}{(x-2)}=\frac{4(x-1)(x+2)(x-2)}{(x-2)}$

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

Hence the correct option is B.

Answer 2

Answer:

It's B

Step-by-step explanation: