Question on Factoring

Question

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

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

Answer 1

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

Answer 2

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:

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

Answer 3

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.

Answer 4

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

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