I need help factoring this polynomial.

Question

I am a competition math student, and I am learning how to factor polynomials. I am trying to factor $x^4-7x^3+4x^2+7x-4$ . I am specifically learning to factor by grouping, but when I try it on this, I get $x^4+4(x^2-1)-7x(x^2-1)$, which further simplifies to $x^4+(x^2-1)(4-7x)$ . I am not sure what to do after this and would appreciate any help! Thanks!

Answer 1

Step 1: Look for linear factors. The key here is the rational root theorem; $\frac{b}{a}$ can be a root and $ax-b$ can be a factor only if $a$ divides the coefficient of the largest power of $x$ and $b$ divides the constant term.

Applying that to our polynomial $p$ here, we have possible rational roots $1,-1,2,-2,4,-4$. Testing... $p(1)=1$, $p(-1)=1$, $p(2)=16-56+16+14-4=-14$, $p(-2)=16+56+16-14-4=70$, $p(4)=256-448+64+28-4=-104$, and $p(-4)=256+448+64-28-4=736$. No rational roots here.

Since this is a fourth-degree polynomial, that's not the end. Quadratic factors are possible - we might be able to write $p(x)=q_1(x)q_2(x)=(x^2+a_1x+b_1)(x^2+a_2x+b_2)$. Now, a little trick - we have a few small values of $p(x)$ to restrict the possibilities. From $p(0)=4$, we get that $b_1b_2=4$, leaving a small list of possibilities. Even stronger, $p(1)=p(-1)=1$, so either $q_1(1)=q_2(1)=1$ or $q_1(1)=q_2(1)=-1$ and similarly at $-1$.

We have $q_1(1)-q_1(-1)=2a_1$, so if they're equal $q_1$ is even - and so is $q_2$, and so is their product. Their product isn't even, so this doesn't happen. That leaves two cases:

• $q_1(1)=q_2(1)=1,q_1(-1)=q_2(-1)=-1$. The polynomials are $x^2+x+b_1$ and $x^2+x+b_2$. Their product is $x^4+2x^3+\text{stuff}$ - nope.
• $q_1(1)=q_2(1)=-1, q_1(-1)=q_2(-1)=1$. The polynomials are $x^2-x+b_1$ and $x^2-x+b_2$. Their product is $x^4-2x^3+\text{stuff}$ - nope.

So, then, the polynomial you were looking at is irreducible over the rationals. No wonder you weren't getting anywhere.