# Can this be factored?

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Can $y^2-2xy-3x$ be factored?

I tried to do it:

$y^2-(2y-3)x$

$\rightarrow$$y(y-3)x-1(y-3)x$ and that finally gave me

$(y-3)x(1+y)$ Is this correct?

Edited by Chikis
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No it cannot, since if you multiply $(y-3)x(1+y)$ it would be: $y^2x-2xy-3x$
If you wanted to factor it would be
$y^2-x(2y+3)$

OR

$y(y-2x)-3x$

but that would be pretty much the end of it. In my eyes at least, it may be wrong

Edited by bluescience
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So nothing can be done again?

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So nothing can be done again?

What do you want to do?

You can write y as a function of x for example. You can construct the general solution for the equation you have given.

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Just take look at this expression, am required to reduce it to its lowest term.

$\frac{3(x^2-y^2)}{y^2-2xy-3x}$

Factoring the numerator, we get

$\frac{3(x+y)(x+y)}{y^2-2xy-3x)}$, but doing a perfect factoring at the denominator to see how terms could cancel out becomes a major problem. So how does one go about this?

Edited by Chikis
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You have a minus sign wrong. You need the difference of two squares formula.

I am not sure what the simplest way of writing your expression is. Do you have a target expression given to you?

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You have a minus sign wrong.

Which minus sign are you talking about here?
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In the difference of two squares...

Expand out (x+y)(x+y) and then (x+y)(x-y) you see that the minus sign makes a big difference.

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In the difference of two squares...Expand out (x+y)(x+y) and then (x+y)(x-y) you see that the minus sign makes a big difference.

Oh! Sorry! I made a mistake there. I intended writing $\frac{3(x+y)(x-y)}{y^2-2x-3x}$ before I typed that trash. Now that the mistake has been corrected, can the denominator be factored out completely to see the terms that could cancel out?

Edited by Chikis
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$\frac{3(x+y)(x-y)}{y^2-2xy-3x}$

Well the solution to the problem above is $\frac{(y-x)}{(x-\frac{y}{3})}$. The issue now is that I cannot give any justification to the solution. Any help here?

Edited by Chikis

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