# Thread: [Factorization] Factorization of polynomials

1. ## [Factorization] Factorization of polynomials

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.

2. 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.

3. 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.

4. 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.

5. Oh, I see. Thanks.

6. 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:
$x^4+5x^3+2x^2+x-3=0$

7. So does anyone have a method to solve this?

8. 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?

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

10. 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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