Which divisor of x3 + 3x2 - 16x - 48 has a remainder of zero? A. X + 2
B. X-4
C. x-3
D. x + 8
I need this one ASAP

Explanation:

We'll use the remainder theorem.

This theorem says that if you divide p(x) over (x-k), then the remainder is p(k).

We have p(x) = x^3 + 3x^2 - 16x - 48 and the denominator is one of the four choices A through D.

Let's go through each of them until we find the answer.

For choice A, we have x+2 which is the same as x-(-2). It matches with x-k to show that k = -2.

Plug k = -2 into p(x) to get

p(x) = x^3 + 3x^2 - 16x - 48

p(-2) = (-2)^3 + 3(-2)^2 - 16(-2) - 48

p(-2) = -12

The remainder is -12, which is not 0 like we're after. We can cross A off the list.

We'll follow the same ideas for choices B through D

For choice B, we plug in k = 4 because x-4 matches with x-k to show that k = 4.

p(x) = x^3 + 3x^2 - 16x - 48

p(4) = (4)^3 + 3(4)^2 - 16(4) - 48

p(4) = 0

We end up with a remainder of 0, so we can stop here. Dividing (x^3+3x^2-16x-48) all over (x-4) leads to some quotient and a remainder of 0.

Put another way: (x-4) is a factor of x^3+3x^2-16x-48

Side notes: