I have been stuck on factorizing this:


I thought I could solve it by making $(a+b)$ as one factor but it didn't work then I tried to add and deduct some terms which that didn't lead me to anything either.

I don't really know what to do next.

  • $\begingroup$ $a^2-2ab+a^2b-2b^2$ does not have any simple factors. $a^2-2ab+ab-2b^2=a^2-ab-2b^2$ does, and $a^2-2ab+a^2b-2ab^2$ does too $\endgroup$
    – Henry
    Jul 31, 2020 at 23:13
  • $\begingroup$ it is a cubic, non-homogeneous polynomial. $\endgroup$
    – Will Jagy
    Jul 31, 2020 at 23:33
  • $\begingroup$ It is quadratic in either $a$ or $b$ separately, so at worst the quadratic formula gives a factorization, albeit involving radicals in terms of the other variable... $\endgroup$ Jul 31, 2020 at 23:36
  • $\begingroup$ @paulgarrett, How about if we set the expression to $0$ and draw the graph? It shows that it is the product of two expressions!! $\endgroup$
    – Peter
    Jul 31, 2020 at 23:40
  • $\begingroup$ Here's what Wolfram Alpha has to say about the problem. Doesn't look like there's a simple factorisation, as Henry pointed out. Not sure what you mean by setting it to 0 and drawing a graph, either. $\endgroup$
    – doobdood
    Jul 31, 2020 at 23:44

1 Answer 1




In order to factorize the last polynomial, we have to solve the following quadratic equation:


that is equivalent to



$b=\frac{a(a-2)\pm\sqrt{a^2(a^2-4a+12)}}{4}=\frac{a(a-2)\pm a\sqrt{a^2-4a+12}}{4}.$

Now we can factorize the polynomial:



$\begin{align*} &=-\frac{1}{8}\left(4b-a^2+2a+a\sqrt{a^2-4a+12}\right)\cdot\left(4b-a^2+2a-a\sqrt{a^2-4a+12}\right).\\ \end{align*}$

Therefore we get that

$\begin{align} &a^2-2ab+a^2b-2b^2=\\ &=-\frac{1}{8}\left(4b-a^2+2a+a\sqrt{a^2-4a+12}\right)\cdot\left(4b-a^2+2a-a\sqrt{a^2-4a+12}\right).\\ \end{align}$

  • 1
    $\begingroup$ I give to you also a credit for the recent comment in an answer. $\endgroup$
    – Sebastiano
    Aug 22, 2020 at 21:15

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