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Can this be factored?


Chikis
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Can [math]y^2-2xy-3x[/math] be factored?

I tried to do it:

[math]y^2-(2y-3)x[/math]

[math]\rightarrow[/math][math]y(y-3)x-1(y-3)x[/math] and that finally gave me

[math](y-3)x(1+y)[/math] Is this correct?

Edited by Chikis
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No it cannot, since if you multiply [latex](y-3)x(1+y)[/latex] it would be: [latex]y^2x-2xy-3x[/latex]
If you wanted to factor
8bf5ae11bdf5aee694048e56ae33a2ae-1.png
it would be
[latex]y^2-x(2y+3)[/latex]

OR

[latex]y(y-2x)-3x[/latex]

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?

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.

[math]\frac{3(x^2-y^2)}{y^2-2xy-3x}[/math]

Factoring the numerator, we get

[math]\frac{3(x+y)(x+y)}{y^2-2xy-3x)}[/math], 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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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 [math]\frac{3(x+y)(x-y)}{y^2-2x-3x}[/math] 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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[math]\frac{3(x+y)(x-y)}{y^2-2xy-3x}[/math]

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

Edited by Chikis
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