Factoring a third degree polynomial as part of a limits problem

[EcKoh]

Homework Statement

I have a limit problem, however I do know how to work limits, I guess what I need is more of a refresher on how to work third degree polynomials. The polynomial(s) I am trying to work with are the following:

x³-2x²+2x-15
-and-
x³-5x²+10x-12

The limit is a limit where x approaches 3, so I believe that it is correct to assume that 3 is the rational root that we use in this case. However, my attempts haven't come up with a correct solution. I am just wondering where I have gone wrong and what I need to do to reach the correct answer.

2. The attempt at a solution

x³-2x²+2x-15
x³-2x²+2x-15 ÷ x+3

First, I divide to get x², then multiply x-3 by this to reach x³-3x²
Then I bring down 2x to get x²+2s
Next I divide again to get x, then multiply x-3 again to reach x²-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?

 

[LCKurtz]

Homework Statement

I have a limit problem, however I do know how to work limits, I guess what I need is more of a refresher on how to work third degree polynomials. The polynomial(s) I am trying to work with are the following:

x³-2x²+2x-15
-and-
x³-5x²+10x-12

The limit is a limit where x approaches 3, so I believe that it is correct to assume that 3 is the rational root that we use in this case. However, my attempts haven't come up with a correct solution. I am just wondering where I have gone wrong and what I need to do to reach the correct answer.

2. The attempt at a solution

x³-2x²+2x-15
x³-2x²+2x-15 ÷ x+3

First, I divide to get x², then multiply x-3 by this to reach x³-3x²
Then I bring down 2x to get x²+2s
Next I divide again to get x, then multiply x-3 again to reach x²-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?

A root x = 3 corresponds to a factor of (x-3). Try that.

 

[vela]

First, I divide to get x², then multiply x-3 by this to reach x³-3x²
Then I bring down 2x to get x²+2s
Next I divide again to get x, then multiply x-3 again to reach x²-3x
Finally I bring down -15 to get x-15. However if I divide again I get 1, but 1•-3 does not equal -15 so I get a remainder of 12.

What am I doing wrong here?

When you subtracted $x^2-3x$ from $x^2+2x$, you made a sign error.

 

[dextercioby]

If you can't get a hold on what <polynomial long division> means, then try this thing

$$x^3 - 2x^2 + 2x -15 = (x-3) (Ax^2 + Bx + C)$$

No matter what value of x. Then try x=0, x=1, x=-1...

 

[e^(i Pi)+1=0]

edit - never mind

 

[EcKoh]

Thanks to all of you. I realize now what my errors were.