1. ## Polynomial factoring

Hey guys.. having a problem with this.. I feel like the step I am missing is pretty basic, so you should be able to spot it quickly.. I am having a problem with my long division.

Here it is in paint:

Polynomial.jpg picture by AKelsoY - Photobucket

2. on your last two lines ...

x^2 - 1
x^2 + x
--------
-x - 1

now ... x+1 will go into -x-1 how many times?

3. -1?

So it will be 2x^2+x-1?

4. Originally Posted by Slipery
-1?

So it will be 2x^2+x-1?
you check it ...

is $\displaystyle (x+1)(2x^2+x-1) = 2x^3 + 3x^2 - 1$ ?

5. Yup! thanks

6. I got it..

I am having another problem though..

I cannot factor (2x^2 - 1)??? I have tried many combinations.. I have been trying this for 20 minutes, so its a bit frustrating

I hope its not something as easy as leaving the two in front of the bracket. This is for a question where I have to solve the equation. How would having a 2 in front of the bracket affect my zero?..

The question is 'Solve 2x^3+6x^2-x-3= 0'
So I have (x+3)(2x^2-1)

Now what? =(

7. if you solve $\displaystyle 2x^2 - 1$ using the quadratic formula:

$\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

the roots come out to be:

$\displaystyle (x + \frac{\sqrt{2}}{2})$ & $\displaystyle (x - \frac{\sqrt{2}}{2})$

Hope this helps