# How to factorize $a^2-2ab+a^2b-2b^2$?

I have been stuck on factorizing this:

$$a^2-2ab+a^2b-2b^2$$

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.

• $a^2-2ab+a^2b-2b^2$ 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 Jul 31, 2020 at 23:13
• it is a cubic, non-homogeneous polynomial. Jul 31, 2020 at 23:33
• 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... Jul 31, 2020 at 23:36
• @paulgarrett, How about if we set the expression to $0$ and draw the graph? It shows that it is the product of two expressions!! Jul 31, 2020 at 23:40
• 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. Jul 31, 2020 at 23:44

$$a^2-2ab+a^2b-2b^2=$$

$$=-2b^2+a(a-2)b+a^2.$$

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

$$-2b^2+a(a-2)b+a^2=0$$

that is equivalent to

$$2b^2-a(a-2)b-a^2=0.$$

$$\Delta=a^2(a-2)^2+8a^2=a^2(a^2-4a+12)\ge0,$$

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

Now we can factorize the polynomial:

$$-2b^2+a(a-2)b+a^2=$$

$$=-2\left(b-\frac{a(a-2)-a\sqrt{a^2-4a+12}}{4}\right)\cdot\left(b-\frac{a(a-2)+a\sqrt{a^2-4a+12}}{4}\right)=$$

\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}

• I give to you also a credit for the recent comment in an answer. Aug 22, 2020 at 21:15