# Factorization of the polynomial

• Nov 30th 2010, 07:59 PM
Aero763
Factorization of the polynomial
I need to factor the next polynomial:

$
3x^3 + 7x^2 - 9x + 2
$
• Nov 30th 2010, 08:50 PM
Soroban
Hello, Aero763!

Quote:

$\text{Factor: }\:3x^3 + 7x^2 - 9x + 2$

Using the Rational Roots Theorem, the only candidates are: . $\pm1,\:\pm2,\:\pm\frac{1}{3},\:\pm\frac{2}{3}$

We find that $x \,=\,\frac{2}{3}$ is a root.
. . Hence, $(3x-2)$ is a factor.

Long Division: . $3x^3 + 7x^2 - 9x + 2 \;=\;(3x-2)(x^2 + 3x - 1)$

• Nov 30th 2010, 08:55 PM
Grep
That one took a bit of work. First, do you know how to do synthetic division? I started with possible roots:
$\frac{\pm{1}~\pm{2}}{\pm{1}~\pm{3}}$
I mean that $\pm{1}, \pm{2}, \pm{\frac{1}{3} and \pm{\frac{2}{3}$ are possible roots to start with. Then I tried them with synthetic division to find one with a remainder of 0. Know how to do that?

I'm thinking there's a more elegant way than to find possible roots and try them one by one, but it does work. If someone knows that more elegant way, please do tell! It's been a while since I've had to do this.

EDIT: Drat, Soroban, you beat me to it! (Tongueout)
ANOTHER EDIT: The Rational Roots Theorem is explained here, if that helps: Rational root theorem - Wikipedia, the free encyclopedia
• Nov 30th 2010, 10:47 PM
CaptainBlack
Quote:

Originally Posted by Grep
That one took a bit of work. First, do you know how to do synthetic division? I started with possible roots:
$\frac{\pm{1}~\pm{2}}{\pm{1}~\pm{3}}$
I mean that $\pm{1}, \pm{2}, \pm{\frac{1}{3} and \pm{\frac{2}{3}$ are possible roots to start with. Then I tried them with synthetic division to find one with a remainder of 0. Know how to do that?

Why not substitute the candidate into the polynomial and see if you get 0.

CB
• Dec 1st 2010, 11:24 AM
Grep
Quote:

Originally Posted by CaptainBlack
Why not substitute the candidate into the polynomial and see if you get 0.

Good point, that's certainly easier to explain than synthetic division in a forum without drawings.

On the other hand, bet if I raced you to see if 2/3 is a root, and I used synthetic division, and you substituted into the polynomial, that I'd finish first. Simpler operations, and less of them. Not to mention that you would still have to perform the division to find the quotient once you find a root.

You'd have, if I'm not mistaken, one cubing, one squaring, 3 multiplications and 4 additions/subtractions.

I'd have 3 multiplications and 3 additions (or subtractions, depending on your method).

That aside, it's good that you point out that alternate method. I think it helps to see what the roots really mean, and what their properties are.
• Dec 1st 2010, 07:13 PM
Aero763
Thank to your help, I learned the algorithm of factoring this type of polynomials. Couldn't do it without your help in a million years. Thank you guys!

In case somebody else is gonna try to solve it, there's some videos on the topic.