How do you find the product #(2a-9)(3a^2+4a-4)#?

2 Answers
May 12, 2018

Answer:

#6a^3-19a^2-44a+36#

Explanation:

#(2a-9)(3a^2+4a-4)#

#=2a(3a^2+4a-4)-9(3a^2+4a-4)#

#=(6a^3+8a^2-8a)-(27a^2+36a-36)#

#=6a^3+8a^2-8a-27a^2-36a+36#

#=6a^3+8a^2-27a^2-8a-36a+36#

#=6a^3-19a^2-44a+36#

May 12, 2018

Answer:

Your product is #6a^3-19a^2-44a+36#

Explanation:

To find the product you have to multiply each term in the one parenthesis with each term in the second parenthesis and add together.

In this instance you will first multiply all the terms in #(3a^2+4a-4)# with #2a#, thereafter all the terms in #(3a^2+4a-4)# with 9 and then add together

  1. #(3a^2+4a-4)*2a=6a^3+8a^2-8a#
  2. #(3a^2+4a-4)*9=27a^2+36a-36#

But as 9 is negative, "adding together" implies that the 2nd expression must be subtracted from the 1st:
#(6a^3+8a^2-8a)-(27a^2+36a-36)#
#=6a^3+(8-27)a^2-(8+36)a+36#
(remember that #-(-36)=+36#)
#=6a^3-19a^2-44a+36#