How do you factor completely #16x^2-81#?

1 Answer
Jan 13, 2017

Answer:

See full factoring process below:

Explanation:

Because there is no #x# term and the constant term is a negative we know the sum of the factored #x# terms is #0#. And, because both the #x^2# coefficient and the constant are squares this expression can be factored as:

#(ax + b)(ax - b)# or

#(4x + 9)(4x - 9)#