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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

Best Answer 

 #1
avatar+2268 
+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
 #1
avatar+2268 
+1
Best Answer

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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