# Factoring linear factors.

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• Apr 28th 2013, 02:00 PM
FatimaA
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.
• Apr 28th 2013, 02:23 PM
topsquark
Re: Factoring linear factors.
Quote:

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 $x - 2$ divides $x^3 - 2x^2 + 8x - 16$

Quote:

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
• Apr 28th 2013, 02:26 PM
HallsofIvy
Re: Factoring linear factors.
Quote:

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 $x^3- 2x^2= x^2(x- 2)$ exactly " $x^2$" 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.

Quote:

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.

Quote:

Thanks in advance for any help I can get on this.
• Apr 28th 2013, 02:35 PM
FatimaA
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))