Chikis Posted April 18, 2016 Share Posted April 18, 2016 (edited) 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 April 18, 2016 by Chikis Link to comment Share on other sites More sharing options...

bluescience Posted April 18, 2016 Share Posted April 18, 2016 (edited) 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 factorit 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 April 18, 2016 by bluescience Link to comment Share on other sites More sharing options...

Chikis Posted April 18, 2016 Author Share Posted April 18, 2016 So nothing can be done again? Link to comment Share on other sites More sharing options...

ajb Posted April 18, 2016 Share Posted April 18, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 19, 2016 Author Share Posted April 19, 2016 (edited) 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 April 19, 2016 by Chikis Link to comment Share on other sites More sharing options...

ajb Posted April 19, 2016 Share Posted April 19, 2016 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? Link to comment Share on other sites More sharing options...

Chikis Posted April 19, 2016 Author Share Posted April 19, 2016 You have a minus sign wrong. Which minus sign are you talking about here? Link to comment Share on other sites More sharing options...

ajb Posted April 19, 2016 Share Posted April 19, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 21, 2016 Author Share Posted April 21, 2016 (edited) 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 April 21, 2016 by Chikis Link to comment Share on other sites More sharing options...

Chikis Posted April 22, 2016 Author Share Posted April 22, 2016 (edited) [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 April 22, 2016 by Chikis Link to comment Share on other sites More sharing options...

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