Question

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

Answer 1

This is simply a difference of squares.

`(a^2-b^2)=(a+b)(a-b)`

Thus

`y = (11x - 1)(11x + 1)`

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!

Answer 2

See a solution process below:

Explanation:

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

`color(red)(a)^2 - color(blue)(b)^2 = (color(red)(a) + color(blue)(b))(color(red)(a) - color(blue)(b))`

Let `color(red)(a)^2 = 121x^2` then `color(red)(a) = sqrt(121x^2) = 11x`

Let `color(blue)(b)^2 = 1` then `color(blue)(1) = sqrt(1) = 1`

Substituting gives:

`color(red)(121x^2) - color(blue)(1)^2 => (color(red)(11x) + color(blue)(1))(color(red)(11x) - color(blue)(1))`

We can then write the equation factored as:

`y = (11x + 1)(11x - 1)`