[Factorization] Factorization of polynomials

**Cthul** · March 30th 2010, 12:24 AM

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

**HallsofIvy** · March 30th 2010, 01:56 AM

Originally Posted by Cthul

I need help on how to factorize the following w/o calculator if possible.
$x^4+5x^3+2x^2+x-3=0$
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.

**Cthul** · March 30th 2010, 02:11 AM

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

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

**mr fantastic** · March 30th 2010, 02:58 AM

Originally Posted by Cthul

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

**Cthul** · March 30th 2010, 03:16 AM

Oh, I see. Thanks.

**Cthul** · April 1st 2010, 06:28 PM

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?
$(x^2+ax+b)(x^2+cx+d)=0$
Expanded.
$x^4+ax^3+cx^3+acx^2+bx^2+dx^2+adx+bcx+bd=0$
Common Factor.
$x^4+x^3(a+c)+x^2(ac+b+d)+x(ad+bc)+bd=0$

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

Original Equation:
$x^4+5x^3+2x^2+x-3=0$

**Cthul** · April 8th 2010, 11:18 PM

So does anyone have a method to solve this?

**mr fantastic** · April 8th 2010, 11:57 PM

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?

**Cthul** · April 8th 2010, 11:58 PM

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

**Bacterius** · April 9th 2010, 12:15 AM

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 $a, b, c$ and $d$ , 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 ?

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