dividing polynomial

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March 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?
March 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.
March 16th 2009, 01:08 PM
allywallyrus

oh, sorry about that :)
thanks for the quick response, we never went over synthetic division so i've been doing it the long way.
much appreciated!
March 16th 2009, 01:09 PM
masters

Quote:

Originally Posted by allywallyrus

oh, sorry about that :)
thanks for the quick response, we never went over synthetic division so i've been doing it the long way.
much appreciated!

You get the same quadratic polynomial if you use long division.