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

2 Answers
Mar 7, 2018

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!

Mar 7, 2018

Answer:

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)#