# How do you factor y= 121x^2 - 1?

Mar 7, 2018

This is simply a difference of squares.

$\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$

Thus

$y = \left(11 x - 1\right) \left(11 x + 1\right)$

We can immediately factor it like this because we see that there is no $x$ term, and the ${x}^{2}$ and ${x}^{0}$ terms are perfect squares.

Hopefully this helps!

Mar 7, 2018

See a solution process below:

#### Explanation:

The right side of the equation is a special form of quadratic:

${\textcolor{red}{a}}^{2} - {\textcolor{b l u e}{b}}^{2} = \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right)$

Let ${\textcolor{red}{a}}^{2} = 121 {x}^{2}$ then $\textcolor{red}{a} = \sqrt{121 {x}^{2}} = 11 x$

Let ${\textcolor{b l u e}{b}}^{2} = 1$ then $\textcolor{b l u e}{1} = \sqrt{1} = 1$

Substituting gives:

$\textcolor{red}{121 {x}^{2}} - {\textcolor{b l u e}{1}}^{2} \implies \left(\textcolor{red}{11 x} + \textcolor{b l u e}{1}\right) \left(\textcolor{red}{11 x} - \textcolor{b l u e}{1}\right)$

We can then write the equation factored as:

$y = \left(11 x + 1\right) \left(11 x - 1\right)$