# Factoring 3rd degree polynomials

• Dec 5th 2010, 01:52 PM
l flipboi l
Factoring 3rd degree polynomials
Hello,

Can you please show me how to factor this out x^3 - 6x^2 + 11x - 6?

I would like to see how its done, step-by-step if possible.

Thanks!
• Dec 5th 2010, 01:59 PM
Jerry99
Id start out by finding a zero on my cacluator. For example, 2 is a zero in that function. And, then Id do synthetic division to make it quadratic and then factor it. Are you familier with synthetic division?
• Dec 5th 2010, 02:03 PM
rtblue
Well, from intuition we can gather that x=1 is a root. Now we have:

$\displaystyle (x-1)(x^2-5x+6)=0$

$\displaystyle (x-1)(x-3)(x-2)=0$

$\displaystyle x=1,x=3,x=2$
• Dec 5th 2010, 02:10 PM
l flipboi l
Thanks! i just figured it out too by using synthetic division.

I guess my question now is, would plugging in numbers to find it equal to zero be the best approach when dealing with 3rd degree polynomials or higher?
• Dec 5th 2010, 05:37 PM
rtblue
There are many approaches to solving polynomials of a high degree. Synthetic division is probably the best one. Lets call the leading coefficient of an n degree polynomial Q and lets call the last term of the same polynomial P. Now, calculate all the values of P/Q and use synthetic division, with those values (only those values could possibly work). Have you heard of Descarte's Rule of Signs? It gives you some direction as to where the roots might be.

Descartes' Rule of Signs
• Dec 6th 2010, 04:05 AM
HallsofIvy
Quote:

Originally Posted by l flipboi l
Thanks! i just figured it out too by using synthetic division.

I guess my question now is, would plugging in numbers to find it equal to zero be the best approach when dealing with 3rd degree polynomials or higher?

That depends upon what you mean by "plugging in numbers". You have to choose those numbers carefully. When rtblue said "from intuition we can gather that x=1 is a root" I suspect he meant that it is easy to see that 1- 6+ 11- 6= 12- 12= 0. You can also use the "rational root theorem"- if the leading coefficient is a and the constant term of a polynomial is b, then any rational root if of the form m/n where n divides a and m divides b. Here, the leading coefficient is 1 and the constant term is 6 so that the only possible rational roots are the integers $\displaystyle \pm 1$, $\displaystyle \pm 2$, $\displaystyle \pm 3$, and $\displaystyle \pm 6$. If NONE of those turned out to be a root, there would be no way to factor this with integer coefficients.
• Dec 6th 2010, 12:24 PM
l flipboi l
Thanks! i'm gonna read the rules of signs.