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

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Answer 1 · 2018-03-07T00:01:47

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 · 2018-03-07T00:03:40

See a solution process below:

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:

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

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