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.
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.
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:
$\displaystyle x^4+5x^3+2x^2+x-3=0$
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 ?