I have very basic Question about factoring, we know that,

$$x^2+2xy+y^2 = (x+y)^2$$ $$x^2-2xy+y^2 = (x-y)^2$$

But what will

$$x^2-2xy-y^2 = ??$$ $$x^2+2xy-y^2 = ??$$

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    $\begingroup$ Your second line should be $x^2-2xy+y^2$. $\endgroup$ – Jam Aug 11 '14 at 19:43
  • $\begingroup$ I corrected the equation in the above question $\endgroup$ – Shoaibkhanz Aug 11 '14 at 19:47
  • $\begingroup$ @Shoaibkhanz I like your question - it shows that you are thinking about what you know in a good way, and that will serve you well as you learn more. You should look carefully at the way that Thomas Andrews has used the other common identity $x^2-y^2=(x+y)(x-y)$ which works whenever you have the difference of two squares (or of two positive expressions even) - then you will really start to understand what is going on - even if it takes you a little time, it will be time well spent. $\endgroup$ – Mark Bennet Aug 11 '14 at 20:32
  • $\begingroup$ Certainly Mark! , its interesting how he splits that into an identity, in addition to that John provides the Formula way of achieving it $\endgroup$ – Shoaibkhanz Aug 12 '14 at 17:07

You can use the quadratic formula:

$$x^2 + (-2y)x + (-y^2) = 0\to \\ x= \frac{2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$

So $x^2 - 2xy - y^2 = [x - (1+\sqrt{2})y][x - (1-\sqrt{2})y].$

For the other case,

$$x^2 + (2y)x + (-y^2) = 0\to \\ x= \frac{-2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =-y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$

So $x^2 + 2xy - y^2 = [x - (1-\sqrt{2})y][x - (1+\sqrt{2})y].$


There is no simple factorization of $x^2+2xy-y^2$ nor $x^2-2xy-y^2$, although you can write:

$$\begin{align}x^2+2xy-y^2 &= (x+y)^2-2y^2 \\&= \left(x+y(1+\sqrt{2})\right)\left(x+y(1-\sqrt{2})\right) \end{align}$$

and similarly:

$$x^2-2xy-y^2 = \left(x-y(1+\sqrt{2})\right)\left(x-y(1-\sqrt{2})\right)$$


This is another way of solving this problem using Completing the Square method. $x^2+2xy-y^2=??$

First we have,



By adding both sides by $y^2$ to make it perfect square then,




$x=(+ or -)[2^(1/2)]y-y$

$x=[2^(1/2)]y-y ; x=-[2^(1/2)]y-y $

$x=[2^(1/2)-1]y ; x=-[2^(1/2)-1]y$

therefore we have, ${x-[2^(1/2)-1]y}{x+[2^(1/2)-1]y}$

By solving $x^2-2xy-y^2=??$ Try to solve this by relying with this pattern. Thank you.

  • $\begingroup$ I don't think the answer is complete. Also please check the formatting $\endgroup$ – Shailesh Mar 31 '16 at 11:01

You can factor a quadratic trinomial


by finding the roots of

$$\frac{ax^2+bxy+cy^2}{y^2}=a\left(\frac xy\right)^2+b\left(\frac xy\right)+c=0.$$


$$ax^2+bx+c=y^2a\left(\frac xy-r_0\right)\left(\frac xy-r_1\right)=a\left(x-r_0y\right)\left(x-r_1y\right).$$


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