Factorize 5 term polynomial with 2 variable. What method is the best?

Question

I can't factorize this polynomial:$$ a^{4}-a^{3}b-7a^{2}b^{2}+ab^{3}+6b^{4}$$ I can't group it. And it have 2 variable. Help me!
Thank you.

Answer 1

It is a homogeneous polynomial, so you can de-homogenise it, try to factot thz resulting polynomial in one variable and re-homogenise the factors.

Set $t=\frac ab$. The polynomial can be rewritten as $$a^{4}-a^{3}b-7a^{2}b^{2}+ab^{3}+6b^{4}=b^4(t^4-t^3-7t^2+t+6).$$ Now, use the rational roots theorem: if the second factor has a rational root, it is a divisor of $6$. It is easy to see that $1$ and $-1$ are indeed roots, and performing the division we obtain $$t^4-t^3-7t^2+t+6=(t^2-1)(t^2-t-6).$$ One also sees that $3$ is a root of the quadratic polynomial $t^2-t-6$, whence the other root is $-2$ (by Vieta's relations) and the final factorisation is $$t^4-t^3-7t^2+t+6=(t-1)(t+1)(t+2)(t-3),$$ whence, teplacing $t$ and multiplying by $b^4$ $$a^{4}-a^{3}b-7a^{2}b^{2}+ab^{3}+6b^{4}=(a-b)(a+b)(a+2b)(a-3b).$$

Answer 2

It is $$\left( -b+a \right) \left( -3\,b+a \right) \left( 2\,b+a \right) \left( b+a \right) $$

Answer 3

\begin{eqnarray} E &=&a^{4}-a^{3}b\underbrace{-a^{2}b^{2}-6a^2b^2}_{-7a^2b^2}+ab^{3}+6b^{4}\\ &=&a^{3}(a-b)-ab^{2}(a-b)-6b^2(a^2-b^{2})\\ &=&(a-b)(a^3-ab^2-6b^2(a+b))\\ &=&(a-b)(a(a^2-b^2)-6b^2(a+b))\\ &=&(a-b)(a+b)(a(a-b)-6b^2)\\ &=&(a-b)(a+b)(a^2-ab-6b^2)\\ &=&(a-b)(a+b)(a-3b)(a+2b) \end{eqnarray}