Re: Factoring Polynomials

Re: Factoring Polynomials

ok thanks!

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

Re: Factoring Polynomials

Re: Factoring Polynomials

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

Re: Factoring Polynomials

Re: Factoring Polynomials

Quote:

Originally Posted by

**nickgc** Hi forum!

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

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

[snip]

Why are you trying to factorise it?

Re: Factoring Polynomials

I am reviewing for our midterm exams...

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

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?

Re: Factoring Polynomials

I'm sorry

Yes, the instruction was to factorise it

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

Re: Factoring Polynomials

Quote:

Originally Posted by

**nickgc** Hi forum!

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

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

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

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.

Re: Factoring Polynomials

Assume that $\displaystyle f=x^5-6x^4-11x^2-12x-16$ can decompose into rational polynomials it will be of form$\displaystyle (ax^2+bx+c)(dx^2+ex+f)(x-q),a,b,c,d,e,f,q\in Q$ so it will have a rational root assume$\displaystyle \frac{p}{q},(p,q)=1$so $\displaystyle (\frac{p}{q})^5-6(\frac{p}{q})^4-11(\frac{p}{q})^2-12\frac{p}{q}-16=0 \Rightarrow p^5-6p^4q-11p^2q^3-12pq^4-16q^5=0 \Rightarrow p|q,q|p\Rightarrow p=q$ contradiction.