# dividing polynomial

• Mar 16th 2009, 12:55 PM
allywallyrus
dividing a polynomial
i'm not sure which forum this belongs in, but this seemed like the closest fit. the question is asking me to solve for x and i'm kinda stuck :)
heres what i have:
$2x^3+6x^2-x=3$
so i brought the 3 over and found out -3 is a factor
$f(-3)=2(-3^3)+6(-3^2)-(-3)-3$
$-54+54+3-3=0$
but then when i divide the initial equation by x+3 it end up with a zero and then i dont really know where to go from there
ideas?
• Mar 16th 2009, 01:05 PM
masters
Quote:

Originally Posted by allywallyrus
i'm not sure which forum this belongs in, but this seemed like the closest fit. the question is asking me to solve for x and i'm kinda stuck :)
heres what i have:
2x^3+6x^2-x=3
so i brought the 3 over and found out -3 is a factor
f(-3)=2(-3^3)+6(-3^2)-(-3)-3
0=-54+54+3-3
but then when i divide the initial equation by x+3 it end up with a zero and then i dont really know where to go from there
ideas?

Hi allywallyrus,

First of all, -3 is not a factor. -3 is a zero (root) of $f(x)=2x^3+6x^2-x-3$

$x+3$ is a factor.

Using synthetic division, we can easily find the depressed polynomial in the 2nd degree.

Code:

``` -3 | 2  6  -1  -3       -6  0  3     -------------     2  0  -1  0```
The depressed polynomial is $2x^2-1=0$, which you can easily find the remaining 2 zeros.
• Mar 16th 2009, 01:08 PM
allywallyrus