Need help to factorize.

[Farzin]

Hi,
I have been stuck on factorizing this: [MATH]a^2-2ab+a^2b-2b^2[/MATH].
I thought I could solve it by making [MATH](a+b)[/MATH] 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 how what to next.
Any help would be greatly appreciated.

 

[Dr.Peterson]

I'm wondering if you might have copied something wrong. With three terms of degree 2 and one of degree 3, it seems like it might not factor, though I can't be sure. One small change would make it easy.

 

[Farzin]

I'm wondering if you might have copied something wrong. With three terms of degree 2 and one of degree 3, it seems like it might not factor, though I can't be sure. One small change would make it easy.

No everything is correct. I know it could be much easier with a small change.

 

[HallsofIvy]

I would be inclined to write \(\displaystyle a^2+ 2ab+ a^2b- 2b^2\) as \(\displaystyle a^2+ 2ab+ b^2+ a^2b- 3b^2= (a+ b)^2+ a^2b- 3b^2= (a+ b)^2+ b(a^2- 3b)\).

That's about the best you can do.

 

[Dr.Peterson]

That's about all I could do, too. But when I set the expression to 0 and graph it, it seems clear that it is the product of two expressions that don't look like polynomials, which is very intriguing. I haven't figured out what they are.



@Farzin, where did this come from?

 

[Farzin]

That's about all I could do, too. But when I set the expression to 0 and graph it, it seems clear that it is the product of two expressions that don't look like polynomials, which is very intriguing. I haven't figured out what they are.

View attachment 20735

@Farzin, where did this come from?

Thanks for your attempts, this was given to me by a friend and we both are working on it.

 

[Otis]

… write \(\displaystyle a^2\) \(\displaystyle +\) \(\displaystyle 2ab+ a^2b- 2b^2\) …

\(\displaystyle (a+ b)^2+ b(a^2- 3b)\)

The term 2ab should be subtracted, making the result

\(\displaystyle (a - b)^2 \,+\, b\hspace{.09em}(a^2 - 3b)\) …

\(\;\)