# Thread: factorizing a 4th degree polynomial

1. ## 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:

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?

2. ## 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

3. ## Re: factorizing a 4th degree polynomial

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.

4. ## 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.

5. ## Re: factorizing a 4th degree polynomial

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.

6. ## Re: factorizing a 4th degree polynomial

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

7. ## Re: factorizing a 4th degree polynomial

very fancy indeed 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.

8. ## 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.