Factor an equation second order in ##x,y## with other variables

[brotherbobby]

Homework Statement

Resolve into two or more factors : $\large{\boldsymbol{2a^2x^2-2(3b-4c)(b-c)y^2+abxy}}$.

Relevant Equations

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

Statement of the problem : Let me copy and paste to the right the problem as it appears in the text.

Attempt : I couldn't go far into the solution. Below is my hopeless attempt.

Request : Any hints would be welcome.

[There are no solutions provided, but the answers are at the back of the text. I don't want to cheat my way out by looking at the factorised answer and then working my way back, though I admit it can lead to some learning too in the process]

 

[Delta2]

Hmm not sure, wolfram does this very fast and it gives 2 factors with 3 terms each that is $$(A+B+C)(D+E+F)$$

 

[brotherbobby]

Thanks but how to go about it? I have the answer already at the back of the text, am sure, but that doesn't help. :-(

 

[Delta2]

Well I don't know, looking at the wolfram answer the terms are +-(ax) +-(by) +-(cy), so I guess something like full expand the expression and then try to identify these terms and/or their squares.

 

[brotherbobby]

The website is good as a check, but not one to learn mathematics from.
I'd like to do it the old way, not least because my text is from the 1800's!

 

[Delta2]

My original idea before looking at wolfram was to write $$(3b-4c)(b-c)y^2=3(b-c)^2y^2-c(b-c)y^2$$ and see if that helps but I abandon it maybe this idea inspires you but I very much doubt it.

 

[brotherbobby]

Why so? Not bad at all. Let me give it a try.
I am clueless with this problem so every little helps.

 

[WWGD]

Homework Statement: Resolve into two or more factors : $\large{\boldsymbol{2a^2x^2-2(3b-4c)(b-c)y^2+abxy}}$.
Relevant Equations: 1. $a^2-b^2=(a+b)(a-b)$

View attachment 339261Statement of the problem : Let me copy and paste to the right the problem as it appears in the text.

Attempt : I couldn't go far into the solution. Below is my hopeless attempt.

View attachment 339259

Request : Any hints would be welcome.

[There are no solutions provided, but the answers are at the back of the text. I don't want to cheat my way out by looking at the factorised answer and then working my way back, though I admit it can lead to some learning too in the process]

Why not multiply back your factored form see if you get back your initial expression?

 

[fresh_42]

What comes to mind is "completing the square". We have ...
\begin{align*}
2a^2x^2-2(3b-4c)(b-c)y^2+abxy&=2a^2x^2-(6b^2-14bc+8c^2)y^2+abxy\\
&=2a^2x^2-(6b^2-14bc+(49/6)c^2)y^2+abxy + (1/6)c^2y^2\\
&=2a^2x^2 -(\sqrt{6}b-(7c/\sqrt{6}))^2 +abxy + (1/6)c^2y^2\\
&\ldots
\end{align*}
... but I admit that I don't have a clue how to go on.

Another idea is to use only the terms with a coefficient $a$ first ...
\begin{align*}
2a^2x^2-2(3b-4c)(b-c)y^2+abxy&=\left(\sqrt{2}ax+\dfrac{by}{2\sqrt{2}}\right)^2-\dfrac{b^2}{8}y^2 \ldots\\
\phantom{2a^2x^2-2(3b-4c)(b-c)y^2+abxy=}&\ldots -\left(\sqrt{6}b-\dfrac{7c}{\sqrt{6}}\right)^2 + \dfrac{c^2}{6}y^2\\
&=\ldots
\end{align*}
... which looks as if $(A-B)(A+B)=A^2-B^2$ could now be applied twice.

This exercise looks like it is meant to practice all variations of the formulas $(A\pm B)(A\mp B)=\ldots$ and completing the square.

 

[Delta2]

Cant see the full solution in wolfram cause I don't have subscription there but it claims as first step to split the product $$4a^2y^2(3b-4c)(b-c) $$into two factors that have sum $aby$. Those AI engines can be amazing but I don't think a human could think that unless of course he was thinking this for like days.

 

[MatinSAR]

Cant see the full solution in wolfram cause I don't have subscription there but it claims as first step to split the product $$4a^2y^2(3b-4c)(b-c) $$into two factors that have sum $aby$. Those AI engines can be amazing but I don't think a human could think that unless of course he was thinking this for like days.

That human didn't need to think. I guess he just expand sth like (ax+b+c)(ax+y-c) to get sth else and asks us to find first two factors. I can't see how it helps to learn sth it's more similar to a puzzle.

I don't know why Latex doesn't work in my posts anymore ...

 

[fresh_42]

I don't know why Latex doesn't work in my posts anymore ...

It requires refreshing the page occasionally.

 

[Delta2]

That human didn't need to think. I guess he just expand sth like (ax+b+c)(ax+y-c) to get sth else and asks us to find first two factors. I can't see how it helps to learn sth it's more similar to a puzzle.

I don't know why Latex doesn't work in my posts anymore ...

as human here you mean the problem setter? I was talking about the guy that tries to solve this problem.

 

[MatinSAR]

It requires refreshing the page occasionally.

Thanks.

as human here you mean the problem setter?

Yes. I misunderstood your idea. Sorry.

 

[pasmith]

Given the coefficient of $x^2$ is $2a^2$, it makes sense to look for a solution $$2(ax + By)(ax + Dy) = 2a^2x^2 + 2a(B + D)xy + 2BDy^2.$$ Since $p = B + D$ and $q = BD$ are known, we can find $B$ and $D$ as the roots of $$z^2 - pz + q = 0.$$

Since those factors involve square roots, we can instead try $$(2ax + By)(ax + Dy) = 2a^2x^2 + a(B + 2D)xy + BDy^2.$$ which leads to $B + 2D = b$ and $BD = -2(3b - 4c)(b - c)$ which by inspection has a solution $B = -(3b - 4c), D = 2(b - c)$.