# factorizing a 4th degree polynomial

• Aug 11th 2013, 03:26 PM
vermillion
factorizing a 4th degree polynomial
hey guys this is probably pretty basic stuff for most of you. im however struggling with this.
i don't have any standard procedure for this and i can't figure out how its done. i just don't know what kind of method i should be searching for.
in my course it was said you are already supposed to know how to do this so im lacking skill on this. here is the problem:
http://i41.tinypic.com/2wr2m2g.jpg
i underlined the equation where the polynomial is factorized. how to do this? whats the method called and how can i learn to do this on any generic 4th degree polynomial?
• Aug 11th 2013, 04:08 PM
chiro
Re: factorizing a 4th degree polynomial
Hey vermillion.

There is a result that allows you to factor general fourth degree polynomial (which you can assume to be true for algebraic problems):

Quartic function - Wikipedia, the free encyclopedia
• Aug 11th 2013, 04:40 PM
vermillion
Re: factorizing a 4th degree polynomial
Quote:

Originally Posted by chiro
Hey vermillion.

There is a result that allows you to factor general fourth degree polynomial (which you can assume to be true for algebraic problems):

Quartic function - Wikipedia, the free encyclopedia

thanks for your response. but i am afraid this article doesn't really help me here. in the article there is a different last term, an e. I bet there is an easier method out there as well. i hope someone can give another method.
and also in this case the a is always 1. i bet there is some kind of fancy polynomial trick for this.
• Aug 11th 2013, 04:46 PM
vermillion
Re: factorizing a 4th degree polynomial
also, chiro, can you maybe if you think that wikipedia article helps, show me how these 2 examples are factorized? i can't figure it out.
• Aug 12th 2013, 01:31 PM
ChessTal
Re: factorizing a 4th degree polynomial
Quote:

Originally Posted by vermillion
also, chiro, can you maybe if you think that wikipedia article helps, show me how these 2 examples are factorized? i can't figure it out.

The \$\displaystyle {x^3} + 4{x^2} + 6x + 4\$ is factorized by finding the roots of \$\displaystyle {x^3} + 4{x^2} + 6x + 4 = 0\$
If it has any integer root it must be ±1 or ±2 or ±4 (actually we can skip the positive solutions since we will have all positive members on the equation so they can't give zero)
Here we see that -2 works and is a root.
Then dividing \$\displaystyle {x^3} + 4{x^2} + 6x + 4\$ with x+2 we get \$\displaystyle {x^2} + 2{x} + 2\$ which has roots -1-i and -1+i.
Ergo its factorization is (x+2)(x+1+i)(x+1-i)

You can use the same method with the other equation also.
It's not a general method. I.e it will not work in most cases. For that you have to use the link chiro gave.
• Aug 12th 2013, 01:41 PM
ChessTal
Re: factorizing a 4th degree polynomial
Quote:

Originally Posted by vermillion
thanks for your response. but i am afraid this article doesn't really help me here. in the article there is a different last term, an e. I bet there is an easier method out there as well. i hope someone can give another method.
and also in this case the a is always 1. i bet there is some kind of fancy polynomial trick for this.

Well there is a "fancy" method for a=1:
The factorization is \$\displaystyle (x - {r_1})(x - {r_2})(x - {r_3})(x - {r_4})\$ where \$\displaystyle {r_1},{r_2},{r_3},{r_4}\$ can be found from here: quartic formula | planetmath.org
• Aug 12th 2013, 04:27 PM
vermillion
Re: factorizing a 4th degree polynomial
very fancy indeed :D ok thanks guys. However i found out i was overcomplicating as apparently in the exams there won't be any cases where you have to solve 4th degree polynomials, in the worst case they are 3th and i can do that. Also i found a note in the lectures that states 4th degrees should be solved numerically but we won't be using calculators during exams so im good.
• Aug 12th 2013, 04:36 PM
chiro
Re: factorizing a 4th degree polynomial
Also you might want to consider the guessing method were you guess a root and then use long division to go from say cubic to quadratic (in which case you use the quadratic formula or further guesses).

So for example if I had say (x-1)(x-2)^2
= (x-1)(x^2 - 4x + 4)
= (x^3 - 4x^2 + 4x - x^2 + 4x - 4)
= x^3 - 5x^2 + 8x - 4

If you tried a test solution of x = 1 you would get zero and thus you could factor (x-1) straight out.