1. ## 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:
$\displaystyle 2x^3+6x^2-x=3$
so i brought the 3 over and found out -3 is a factor
$\displaystyle f(-3)=2(-3^3)+6(-3^2)-(-3)-3$
$\displaystyle -54+54+3-3=0$
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?

2. 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 $\displaystyle f(x)=2x^3+6x^2-x-3$

$\displaystyle x+3$ 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 $\displaystyle 2x^2-1=0$, which you can easily find the remaining 2 zeros.