Factoring a fourth degree polynomial with missing degrees

Can someone explain how to factor this polynomial: $$x^4 - 4x^2 + 9x + 4 = 0.$$

The answer should be this: $$(x^2 - 3x + 4)(x^2 + 3x + 1) = 0,$$ but I can't find a way to figure it out on my own.

If someone could help me with a step by step explanation I would be very grateful.

• The correct term is “depressed quartic”. Jan 7, 2015 at 23:07

You can easily check with the rational root theorem that there are no roots, so we just have to check if there is a factorization into two quadratics. Set $$x^4 - 4x^2 + 9x + 4 = (x^2 + ax + b)(x^2 + cx + d).$$ Comparing coefficients of $x^3$ gives $c = -a$, so we end up with $$x^4 - 4x^2 + 9x + 4 = (x^2 + ax + b)(x^2 - ax + d)$$ where $a,b,d$ satisfy $bd = 4$, $a(d-b) = 9$ and $-a^2 + b + d = 4$. Now the problem is quite easy with some trial and error, for example try all divisors of $4$ for $b,d$.
• I'm assuming he's trying to factor in $\mathbb{Q}$. Otherwise it would be quite hard. Jan 7, 2015 at 19:11