# [Factorization] Factorization of polynomials

• Mar 30th 2010, 12:24 AM
Cthul
[Factorization] Factorization of polynomials
I need help on how to factorize the following w/o calculator if possible.
$\displaystyle x^4+5x^3+2x^2+x-3=0$
I've attempted to factor with whole numbers with no avail.
• Mar 30th 2010, 01:56 AM
HallsofIvy
Quote:

Originally Posted by Cthul
I need help on how to factorize the following w/o calculator if possible.
$\displaystyle x^4+5x^3+2x^2+x-3=0$
I've attempted to factor with whole numbers with no avail.

By the "rational root theorem", the only possible rational roots of that equation would be factors of 3: 1, -1, 3, and -3. Since none of those satisfy the equation, there are no rational roots. The polynomial cannot be factored with integer coefficients.
• Mar 30th 2010, 02:11 AM
Cthul
Okay.
I've checked the answers and the factor of it is.
$\displaystyle (x^2+x+1)(x^2+4x-3)=0$
For the solution to be true
$\displaystyle x=-2 \pm \sqrt {7}$
From what I know, via the answers.
$\displaystyle (x^2+x+1)=0$
Has no solutions because
$\displaystyle \Delta=-3$
$\displaystyle \Delta<0$
Then there are no solutions.

So, is there no way to do this mentally or by hand? Assuming I am factorizing over irrational numbers.
• Mar 30th 2010, 02:58 AM
mr fantastic
Quote:

Originally Posted by Cthul
Okay.
I've checked the answers and the factor of it is.
$\displaystyle (x^2+x+1)(x^2+4x-3)=0$
For the solution to be true
$\displaystyle x=-2 \pm \sqrt {7}$
From what I know, via the answers.
$\displaystyle (x^2+x+1)=0$
Has no solutions because
$\displaystyle \Delta=-3$
$\displaystyle \Delta<0$
Then there are no solutions.

So, is there no way to do this mentally or by hand? Assuming I am factorizing over irrational numbers.

HOI reasonably assumed that you wanted linear factors.

Since linear factors involving integers are not possible, it's natural to then try to factor it as a product of two quadratics:

(x^2 + ax + b)(x^2 + cx + d)

and then see if you can find integer values of a, b, c and d.
• Mar 30th 2010, 03:16 AM
Cthul
Oh, I see. Thanks.
• Apr 1st 2010, 06:28 PM
Cthul
I'm still stuck on this equation. I tried factorizing by identical equations but I end up with too many variables, is there really no way to factor this? I want to know how to factorize this.

(My attempt to factorize)
The identity?
$\displaystyle (x^2+ax+b)(x^2+cx+d)=0$
Expanded.
$\displaystyle x^4+ax^3+cx^3+acx^2+bx^2+dx^2+adx+bcx+bd=0$
Common Factor.
$\displaystyle x^4+x^3(a+c)+x^2(ac+b+d)+x(ad+bc)+bd=0$

So
$\displaystyle 5=a+c$
$\displaystyle 2=ac+b+d$
$\displaystyle 1=ad+bc$
$\displaystyle -3=bd$

Original Equation:
$\displaystyle x^4+5x^3+2x^2+x-3=0$
• Apr 8th 2010, 11:18 PM
Cthul
So does anyone have a method to solve this?
• Apr 8th 2010, 11:57 PM
mr fantastic
Quote:

Originally Posted by Cthul
So does anyone have a method to solve this?

Are you expected to solve that specific sort of quartic equation without a calculator? Has it come from a section of work where using a calculator is not permitted?
• Apr 8th 2010, 11:58 PM
Cthul
It's an exercise from the book, and the section specified that calculators are not to be used.
• Apr 9th 2010, 12:15 AM
Bacterius
Quote:

Originally Posted by Cthul
It's an exercise from the book, and the section specified that calculators are not to be used.

Hello,
you might want to solve the system for $\displaystyle a, b, c$ and $\displaystyle d$, thus recovering the factorized expression of the polynomial and solving it in the standard way.

However I don't know if it is possible to solve this system. You have four unknowns and four equations, but how to solve it I have no idea yet (haven't really looked). Try to work out some algebra around the system and see what you can do ?