# Factoring Polynomials

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• Jul 13th 2011, 01:24 AM
nickgc
Factoring Polynomials
Hi forum!

x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16

what i did is i grouped them:

( x^5 - 6x^4 + 5x^2 ) + ( - 16x^2 - 12x - 16 )

Then i factored each group:

(x^3) ( x - 5 ) ( x - 1 ) + [ -4 ( 4x^2 - 3x - 4 )]

then i don't know what to do next...

i have tried doing trials and errors using synthetic division but i can't solve it.. help me please..

thanks forum
• Jul 13th 2011, 01:28 AM
Prove It
Re: Factoring Polynomials
It doesn't factorise.
• Jul 13th 2011, 01:31 AM
nickgc
Re: Factoring Polynomials
ok thanks!

but how would i know if a polynomial doesn't factorise?
• Jul 13th 2011, 01:32 AM
Prove It
Re: Factoring Polynomials
Use a CAS...
• Jul 13th 2011, 01:36 AM
nickgc
Re: Factoring Polynomials
I'm sorry but i am not familiar with CAS, can you help me with it?
• Jul 13th 2011, 01:37 AM
Prove It
Re: Factoring Polynomials
Quote:

Originally Posted by nickgc
I'm sorry but i am not familiar with CAS, can you help me with it?

A Computer Algebra System - basically, use a calculator if you can't see a simple factor.
• Jul 13th 2011, 01:43 AM
nickgc
Re: Factoring Polynomials
thank you
• Jul 13th 2011, 05:25 AM
mr fantastic
Re: Factoring Polynomials
Quote:

Originally Posted by nickgc
Hi forum!

x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16

[snip]

Why are you trying to factorise it?
• Jul 14th 2011, 11:38 PM
nickgc
Re: Factoring Polynomials
I am reviewing for our midterm exams...
• Jul 15th 2011, 02:48 AM
CaptainBlack
Re: Factoring Polynomials
Quote:

Originally Posted by Prove It
It doesn't factorise.

The fundamental theorem of algebra says it does. But it has no rational roots though it has one real root near 5.696.

CB
• Jul 15th 2011, 02:57 PM
mr fantastic
Re: Factoring Polynomials
Quote:

Originally Posted by nickgc
I am reviewing for our midterm exams...

That does not explain why you're trying to factorise it!

What I was hoping you would explain in repsonse to my question was - Did the question say or suggest you had the factorise it?
• Jul 18th 2011, 05:27 AM
nickgc
Re: Factoring Polynomials
I'm sorry

Yes, the instruction was to factorise it
• Jul 18th 2011, 07:15 AM
CaptainBlack
Re: Factoring Polynomials
Quote:

Originally Posted by CaptainBlack
The fundamental theorem of algebra says it does. But it has no rational roots though it has one real root near 5.696.

CB

Which means that it does not have any "nice" linear factors, but that does not mean that it does not factor into the product of a "nice" quadratic and a "nice" cubic. Though in this case I think not.

CB
• Jul 18th 2011, 10:17 AM
Re: Factoring Polynomials
Quote:

Originally Posted by nickgc
Hi forum!

x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16

Assuming the third term ought to be $5x^3$

if you had a typo on the 4th term also, (+16 instead of -16)
you would have a polynomial which has 3 factors (some repeated)
and 3 different integer roots for f(x) = 0.
• Jul 18th 2011, 12:01 PM
TheodorMunteanu
Re: Factoring Polynomials
Assume that $f=x^5-6x^4-11x^2-12x-16$ can decompose into rational polynomials it will be of form $(ax^2+bx+c)(dx^2+ex+f)(x-q),a,b,c,d,e,f,q\in Q$ so it will have a rational root assume $\frac{p}{q},(p,q)=1$so $(\frac{p}{q})^5-6(\frac{p}{q})^4-11(\frac{p}{q})^2-12\frac{p}{q}-16=0 \Rightarrow p^5-6p^4q-11p^2q^3-12pq^4-16q^5=0 \Rightarrow p|q,q|p\Rightarrow p=q$ contradiction.
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