1. Factoring linear factors.

Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

I got P(2) = 0 but was I supposed to plug in -2?

This is the part I'm stuck on:

b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.

Thanks in advance for any help I can get on this.

2. Re: Factoring linear factors.

Originally Posted by FatimaA
Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

I got P(2) = 0 but was I supposed to plug in -2?
Nope. You got it right. So now you know that $\displaystyle x - 2$ divides $\displaystyle x^3 - 2x^2 + 8x - 16$

Originally Posted by FatimaA
b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.

Thanks in advance for any help I can get on this.
Divide your polynomial by x - 2. Then note that the remaining factor is quadratic. How do you split a quadratic into its factors?

-Dan

3. Re: Factoring linear factors.

Originally Posted by FatimaA
Hi, I'm working through an exam pretest and I came across this problem. Looking through my notes, I can't find examples that show me how to do it. I'm not totally sure what it's asking.

a) Use the Remainder Theorem to find the remainder when x^3-2x^2+8x-16 is divided by x-2.

For this I did P(2) = (2)^3 -2(2)^2+8(2)-16

I got P(2) = 0 but was I supposed to plug in -2?
No, the remainder when a polynomial, p(x), is divided by x- a is p(a) so 2 is correct.
And, in fact, if you checked "the hard way" by actually dividing, you would see that x- 2 divides into $\displaystyle x^3- 2x^2= x^2(x- 2)$ exactly "$\displaystyle x^2$" times, leaving a remainder of 8x- 16= 8(x- 2) and x- 2 obviously divides evenly into that. I don't know why you are questioning you answer.

This is the part I'm stuck on:

b) Use your answer in part a to factor x^3-2x^2+8x-16 into linear factors allowing complex numbers. Clearly explain your reasoning.
You know that x- 2 is one factor. What is the quotient when you divide x^3- 2x^2+ 8x- 16 by x- 2? (Of course, I have given you the answer above.)
The quotient will be quadratic which you can factor once you know the zeros, if necessary by using the quadratic formula.

Thanks in advance for any help I can get on this.

4. Re: Factoring linear factors.

Thanks. I got:

(x^3-2x^2+8x-16)/(x-2) = x^2+8

So the answer is (x-2)(x+2i(square root 2))(x-2i(square root 2))