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Factor (x^2+y^2-z^2)^2-4x^2y^2as the product of four polynomials of degree 1.

Guest Feb 27, 2018

#1
+1807
+1

I have found a way to factor the original expression $$\left(x^2+y^2-z^2\right)^2-4x^2y^2$$.

 $$\left(x^2+y^2-z^2\right)^2-4x^2y^2\\ \left(\textcolor{red}{x^2+y^2-z^2}\right)^2-(\textcolor{blue}{2xy})^2$$ Although this may not be immediately obvious, the original expression is a difference of squares. I rewrote the original expression to make this fact clearer. Remember that$$\textcolor{red}{a}^2-\textcolor{blue}{b}^2=(\textcolor{red}{a}+\textcolor{blue}{b})(\textcolor{red}{a}-\textcolor{blue}{b})$$. Factor the expression as such. I have used colors to make this process easier for your eyes to digest. $$\left(\textcolor{red}{x^2+y^2-z^2}+\textcolor{blue}{2xy}\right)\left(\textcolor{red}{x^2+y^2-z^2}-\textcolor{blue}{2xy}\right)$$ Let's do some rearranging here. $$\left(\textcolor{green}{x^2+2xy+y^2}-z^2\right)\left(\textcolor{green}{x^2-2xy+y^2}-z^2\right)$$ Why am I highlighting this part? Well, both portions happened to be perfect-square trinomials. $$\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)\left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)$$ Do you notice something? Yes, it is another difference of squares. It is time to factor again! $$\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})\\ \left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z})$$ I have factored each one individually. The last step is to combine the factors into one expression. $$(\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z})$$ This expression satisfies the original condition; the final expression is "the product of four polynomials of degree 1."
TheXSquaredFactor  Feb 27, 2018
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#1
+1807
+1
I have found a way to factor the original expression $$\left(x^2+y^2-z^2\right)^2-4x^2y^2$$.
 $$\left(x^2+y^2-z^2\right)^2-4x^2y^2\\ \left(\textcolor{red}{x^2+y^2-z^2}\right)^2-(\textcolor{blue}{2xy})^2$$ Although this may not be immediately obvious, the original expression is a difference of squares. I rewrote the original expression to make this fact clearer. Remember that$$\textcolor{red}{a}^2-\textcolor{blue}{b}^2=(\textcolor{red}{a}+\textcolor{blue}{b})(\textcolor{red}{a}-\textcolor{blue}{b})$$. Factor the expression as such. I have used colors to make this process easier for your eyes to digest. $$\left(\textcolor{red}{x^2+y^2-z^2}+\textcolor{blue}{2xy}\right)\left(\textcolor{red}{x^2+y^2-z^2}-\textcolor{blue}{2xy}\right)$$ Let's do some rearranging here. $$\left(\textcolor{green}{x^2+2xy+y^2}-z^2\right)\left(\textcolor{green}{x^2-2xy+y^2}-z^2\right)$$ Why am I highlighting this part? Well, both portions happened to be perfect-square trinomials. $$\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)\left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)$$ Do you notice something? Yes, it is another difference of squares. It is time to factor again! $$\left([\textcolor{red}{x+y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})\\ \left([\textcolor{red}{x-y}]^2-\textcolor{blue}{z}^2\right)=(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z})$$ I have factored each one individually. The last step is to combine the factors into one expression. $$(\textcolor{red}{x+y}+\textcolor{blue}{z})(\textcolor{red}{x+y}-\textcolor{blue}{z})(\textcolor{red}{x-y}+\textcolor{blue}{z})(\textcolor{red}{x-y}-\textcolor{blue}{z})$$ This expression satisfies the original condition; the final expression is "the product of four polynomials of degree 1."