# How do you factor completely p^2 - 2p +1 - y^ 2 -2yz - z^2?

Jan 3, 2016

So, the equation completely factored is ${\left(p - 1\right)}^{2} - {\left(y + z\right)}^{2}$

#### Explanation:

Pick the squared numbers or constants: ${p}^{2} , 1 \to {1}^{2} = 1 , {y}^{2} , {z}^{2}$.
So, separate the equation in two parts: $\left({p}^{2} - 2 p + 1\right) - \left({y}^{2} + 2 y z + {z}^{2}\right)$. This step is necessary because all squared numbers need to be positive, and as we can see, both in the second part weren't.

In this point, you should go back the steps in the notable products:
"The first squared plus two times the first multiplied by the second plus the second squared":
${\left(p - 1\right)}^{2} = \left({p}^{2} - 2 p + 1\right)$ and ${\left(y + z\right)}^{2} = \left({y}^{2} + 2 y z + {z}^{2}\right)$.

So, the equation completely factored is ${\left(p - 1\right)}^{2} - {\left(y + z\right)}^{2}$