How do you factor completely: #b^2 -7b -44#?

1 Answer
Jul 18, 2015

Answer:

I found:
#(b-11)(b+4)#

Explanation:

I would first try to solve the equation:
#b^2-7b-44=0# using the Quadratic Formula:
#b_(1,2)=(7+-sqrt(49+176))/2=(7+-15)/2=#
#b_1=(7+15)/2=11#
#b_2=(7-15)/2=-4#

So we can use these two results (with opposite signs) and get:
#(b-11)(b+4)=b^2-7b-44#