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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.

Blue
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Peter
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  • $a^2-2ab+a^2b-2b^2$ 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 – Henry Jul 31 '20 at 23:13
  • it is a cubic, non-homogeneous polynomial. – Will Jagy Jul 31 '20 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... – paul garrett Jul 31 '20 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!! – Peter Jul 31 '20 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. – doobdood Jul 31 '20 at 23:44
  • @doobdood, Here is what I mean: https://www.desmos.com/calculator/nf3rqplftw – Peter Jul 31 '20 at 23:52
  • @Peter That just shows that it can be factored (in either $a$ or $b$), but neither of those curves is polynomial itself. For instance you can write it as $(1+b)a^2-2ba-2b^2$ and use the quadratic formula in terms of $a$, but you'll get a non-polynomial expression in terms of $b$. – Steven Stadnicki Aug 01 '20 at 00:39

2 Answers2

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

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

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

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

Angelo
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0

Here is a very simplistic method of factoring $a^2-2ab+a^2b-2b^2$: $$(a-b)^2=a^2-2ab+b^2$$$$\therefore a^2-2ab+ab^2-2b^2=a^2-2ab+b^2-a^2b-3b^2$$$$=(a-b)^2-a^2b-3b^2$$$$=(a-b)^2-b(a^2-3b)$$$$=(a-b)^2-b(a-\sqrt{3b})(a+\sqrt{3b})$$$$\because\text{ for any }a^2-b$$$$a^2-b=(a-\sqrt{b})(a+\sqrt{b})$$$$\therefore a^2-2ab+a^2b-2b^2=(a-b)^2-b(a-\sqrt{3b})(a+\sqrt{3b})$$I very much hope this helps, despite it being a simplistic (and very much high school Algebra like) way of factoring it.

CrSb0001
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