What are the factors of: $$81a^4 +(x-2a)(x-5a)(x-8a)(x-11a)$$
Do I need to solve $(x-2a)(x-5a)(x-8a)(x-11a)$? If yes, is there a better way to do it? I would be grateful if someone offered hints to tackle this problem rather than the whole solution because I really want to solve it myself.
Thank you.
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I would expand all the brackets. Then the coefficient of $a^4$ is $961=31^2$, which hints at the factorization $(x^2\pm 31a^2+cax)(x^2\pm 31a^2+dax)$, $c,d\in\mathbb{R}$. Find $c$ and $d$.
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Hint(s):
You are supposed to exploit the symmetry of the equation. Notice that each bracket on right differs by $3a$, so choosing the right substitution $x \mapsto x+k\cdot a$ (can you deduce what $k$ to use?) will make them align nicely as $$\left(x+\frac{9a}{2}\right)\left(x+\frac{3a}{2}\right)\left(x-\frac{3a}{2}\right)\left(x-\frac{9a}{2}\right).$$ Then notice also that $81a^4=(3a)^4$. Also $c^2-d^2=(c-d)(c+d)$ is quite handy in this problem. Can you use these hints to solve the problem by yourself?
If $x^2-13ax=y,$
We have $(y-22a^2)(y-40a^2)+81a^4=y^2-2y(31a^2)+(31a^2)^2=?$