Let #f(x)=4x +5 and g(x)=4x2 -5#. What is #(f*g) ( x-1)#?

1 Answer
Jun 9, 2018

#16x^3-28x^2-12x-1#

Explanation:

#"evaluate "(f*g)(x)=f(x)xxg(x)#

#(4x+5)(4x^2-5)=16x^3+20x^2-20x-25#

#"to evaluate "(f*g)(x-1)#

#"substitute "x=x-1" into "(f*g)(x)#

#=16(x-1)^3+20(x-1)^2-20(x-1)-25#

#=16(x^3-3x^2+3x-1)+20(x^2-2x+1)-20x+20-25#

#=16x^3-48x^2+48x-16+20x^2-40x+20-20x-5#

#=16x^3-28x^2-12x-1#