# Factoring the polynomial $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$

can you help me factoring the following polynomial?

$$6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$$

I don't know where to start from.

• add and subtract $a^2b^2$ Commented Jan 5, 2018 at 20:31
• Can you think of anything that you could try? Commented Jan 5, 2018 at 20:32

Regroup first,

\begin{align} (6a^4+5a^2b^2-b^4)+(a^3b+ab^3) &= (6a^2-b^2)(a^2+b^2)+ab(a^2+b^2) \\ &= (6a^2+ab-b^2)(a^2+b^2) \\ &= (3a-b)(2a+b)(a^2+b^2) \end{align}

• Thank you. I consider $6a^4 +5a^2b^2 - b^4$ as $6a^4 - a^2b^2 + 6a^2b^2 - b^4$, the next passage is $a^2(6a^2 -b^2) + b^2(6a^2 - b^2)$ and finally $(a^2 +b^2)(6a^2 - b^2)$. A similar concept is applied to $6a^2+ ab - b^2$. Am I right? Commented Jan 6, 2018 at 18:16
• Yes, that's one way. By the way, I'm already so comfort in applying the cross method. Practice makes perfect. Commented Jan 7, 2018 at 9:52
• Okay, I will do more exerices to improve. Thanks a lot! Commented Jan 7, 2018 at 10:21

May be easier to follow if you define $t= a/b$ and, after dividing by $b^4$, write it as: $$6t^4 + t^3 +5t^2 + t - 1$$

At this point you can use the rational root theorem to find two of the roots, and in the end factor it as $(3 t - 1) (2 t + 1) (t^2 + 1)\,$. After that, you can multiply back by $b^4$ to get the form in $a,b$.

This factorises nicely as $$(a^2 + b^2)(3a - b)(2a + b).$$ To find it, one could see that setting $b=-2a$ gives $0$, as well as $b=3a$.

you can rewrite this as $6a^4+6a^2b^2+a^3b+ab^3-b^4-a^2b^2$. Can you finish?

Note that $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$ is homogenous. So set $u=b/a$. Then $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = a^4(6 + u +5u^2 + u^3 - u^4)$.

We try integer roots of $6 + u +5u^2 + u^3 - u^4$ by considering the divisors of $6$. We find $6 + u +5u^2 + u^3 - u^4=-(u + 2) (u - 3) (u^2 + 1)$.

This gives $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = -(2 a + b) (b-3 a) (a^2 + b^2)$.