Factoring a polynomial of $4$th degree

Question

I was wondering how would you factor $$x^4+4x^3+21x^2-20x+25=0\text{ ?}$$ I cannot find a number that allows the expression to equal zero.

Answer 1

One of the universal principles in mathematics is the ability to make a good guess and then eventually prove the guess. Note that $-\omega,-\omega^2$, where $\omega$ is the non-real cube roots of unity, satisfies the equation. This means $x^2-x+1$ is a factor. Using this, the polynomial factors as $$(x^2-x+1)(x^2+5x+25)$$

Answer 2

You really can do this yourself.

A lemma of Gauss says that a polynomial with integer coefficients factors over the integers if it factors over the rationals. in practice, this means there are just four possible products that might work, with integers $A,B$: $$ (x^2 + A x - 5)(x^2 + B x - 5) $$ $$ (x^2 + A x + 5)(x^2 + B x + 5) $$ $$ (x^2 + A x - 1)(x^2 + B x - 25) $$ $$ (x^2 + A x + 1)(x^2 + B x + 25). $$ In all, $A + B = 4$ because of the $4 x^3.$ In the second one, this gives $20 x,$ so the second one is out, but the first one gives $-20x.$ The first then gives $(AB - 10) x^2,$ so $AB = 31. $ As $31$ is prime, we get either $A+B = \pm 32,$ so the first one is out.

In the third we get $(-25 A - B)x,$ so $-25A-B = -20,$ or $25A + B = 20.$ With $A+B = 4,$ we have $24A = 16,$ so $A$ is not an integer.

In the fourth, we get $(25 A + B)x,$ so $25A+B = -20.$ With $A+B = 4,$ we have $24A = -24,$ so $A=-1$ and $B=5.$ We then confirm that $$ (x^2 - x + 1)(x^2 + 5 x + 25) $$ works.