How do you multiply # 2z(3z+2)(z-1)-(z+7)(z-9)#?

Question

1564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-23T19:36:31

Use the distributive property and FOIL and combine like terms to get our answer, #6z^3-3z^2-2z+63#.

Let's split this into two parts. Focusing on the first part, #2z(3z+2)(z-1)#, we can use the distributive property to simplify it:

#2z(3z+2)(z-1)# = #(6z^2+4z)(z-1)#

Next, let's use FOIL to get to its simplest form:

#(6z^2)(z)+(6z^2)(-1)+(4z)(z)+(4z)(-1)#
#6z^3-6z^2+4z^2-4z#
#6z^3-2z^2-4z#

Let's also use FOIL to simplify the other half of our expression:

#(z+7)(z-9)#
#z^2-9z+7z-63#
#z^2-2z-63#

Finally, let's combine the two terms and simplify:

#6z^3-2z^2-4z-(z^2-2z-63)#
#6z^3-2z^2-4z-z^2+2z+63#
#6z^3-2z^2-z^2-4z+2z+63#
#6z^3-3z^2-2z+63#