Factoring Polynomials

**nickgc** · July 13th 2011, 12:24 AM

Hi forum!

I was factoring this polynomial and i got stucked, please help me..

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

**Prove It** · July 13th 2011, 12:28 AM

It doesn't factorise.

**nickgc** · July 13th 2011, 12:31 AM

ok thanks!

but how would i know if a polynomial doesn't factorise?

**Prove It** · July 13th 2011, 12:32 AM

Use a CAS...

**nickgc** · July 13th 2011, 12:36 AM

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

**Prove It** · July 13th 2011, 12:37 AM

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.

**nickgc** · July 13th 2011, 12:43 AM

thank you

**mr fantastic** · July 13th 2011, 04:25 AM

Originally Posted by nickgc

Hi forum!

I was factoring this polynomial and i got stucked, please help me..

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

[snip]

Why are you trying to factorise it?

**nickgc** · July 14th 2011, 10:38 PM

I am reviewing for our midterm exams...

**CaptainBlack** · July 15th 2011, 01:48 AM

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

**mr fantastic** · July 15th 2011, 01:57 PM

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?

**nickgc** · July 18th 2011, 04:27 AM

I'm sorry

Yes, the instruction was to factorise it

**CaptainBlack** · July 18th 2011, 06:15 AM

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

**Archie Meade** · July 18th 2011, 09:17 AM

Originally Posted by nickgc

Hi forum!

I was factoring this polynomial and i got stucked, please help me..

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.

**TheodorMunteanu** · July 18th 2011, 11:01 AM

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.

Thread: Factoring Polynomials

LinkBack

Thread Tools

Display

Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Re: Factoring Polynomials

Similar Math Help Forum Discussions

Factoring polynomials 6 qs

factoring polynomials in Z

Factoring Polynomials and factoring techniques

Factoring Polynomials

Dividing Polynomials AND Factoring Polynomials

Search Tags