Factoring a polynomial expression

Question

I have the polynomial $x^5-10x^4+24x^3+9x^2-33x-12$ which I factorized as follows
$(x^5-10x^4+24x^3)+(9x^2-33x-12)$
$x^3(x^2-10x+24)+3(3x^2-11x-4)$
$x^3(x-6)(x-4)+3(3x^2-11x-4)$. This is how I did it, the expression $3x^2-11x-4$ is irreductible since it satisfies Eisenstein's criterion for p=11. Anyway the answare provided by the book is $(x-4)(x^4-6x^3+9x+3)$.

Answer 1

you can't use Eisenstein because 33 and 10 are coprime i.e. there is no prime that devides 33 and 10.

See here https://en.wikipedia.org/wiki/Polynomial_long_division , in my country long division method is called polynomial division, f.e. $$(x^5 -10x^4 + 24x^3 +9x^2 -33x -12):(x-4)= ...$$

But who to know that to devide by $x-4$? There is a devider-theorem that says that if there is a rational solution it is a devider of $a_0$ , $12$ in your case, so you can try $3,4$ i.e. polynomial division through $x-3,x-4$ and you see the second one works.

Hope, it helps.