2
$\begingroup$

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

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

$\endgroup$
4
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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

First we have,

$x^2+2xy-y^2=0$

$x^2+2xy=y^2,$

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

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

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

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

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

$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.

$\endgroup$
1
  • $\begingroup$ I don't think the answer is complete. Also please check the formatting $\endgroup$ – Shailesh Mar 31 '16 at 11:01
0
$\begingroup$

You can factor a quadratic trinomial

$$ax^2+bxy+cy^2$$

by finding the roots of

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

Then

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.