1. ## 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

2. ## Re: Factoring Polynomials

It doesn't factorise.

3. ## Re: Factoring Polynomials

ok thanks!

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

Use a CAS...

5. ## Re: Factoring Polynomials

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

6. ## Re: Factoring Polynomials

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.

thank you

8. ## Re: Factoring Polynomials

Originally Posted by nickgc
Hi forum!

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

[snip]
Why are you trying to factorise it?

9. ## Re: Factoring Polynomials

I am reviewing for our midterm exams...

10. ## Re: Factoring Polynomials

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

11. ## Re: Factoring Polynomials

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?

12. ## Re: Factoring Polynomials

I'm sorry

Yes, the instruction was to factorise it

13. ## Re: Factoring Polynomials

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

14. ## Re: Factoring Polynomials

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.

15. ## 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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