How do you simplify and restricted value of #(y^2-16)/(y^2+16)#?

1 Answer
Aug 27, 2017

Answer:

See a solution process below:

Explanation:

We can simplify the numerator using this special case for quadratics:

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

#(color(red)(y)^2 - color(blue)(16)^2)/(y^2 + 16) = ((color(red)(y) + color(blue)(4))(color(red)(y) - color(blue)(4)))/(y^2 + 16)#

Because we cannot divide by #0#, the restricted value is:

#y^2 + 16 != 0#

However, because a number squared will always be non-negative (0 or positive), the #y^2# will always be greater than or equal to #0#.

And then, adding #16# to this will give a number greater than or equal to 16.

Therefore, #y^2 + 16# can never be #0# so there are no restrictions.