How do you multiply #(6z^2 - 4z + 1)(8 - 3z)#?

1 Answer
Apr 14, 2016

Answer:

#-18z^3+60z^2-35z+8#

Explanation:

When multiplying polynomials, as we see here, we must distribute everything.

Every term that is in the trinomial #6z^2-4z+1# must be multiplied individually by both terms in the following binomial #8-3z#. Then, all these multiplied terms will be added to one another to form a large polynomial.

Let's break down what we'll multiply:

#"("underbrace(color(green)(6z^2)underbrace(color(blue)(-4z)+underbrace(color(red)1")("8-3z)_(color(red)(1(8-3z))))_color(blue)(-4z(8-3z)))_color(green)(6z^2(8-3z))")"#

So, we see that the #(8-3z)# term is distributed to each term within #(6z^2-4z+1)#.

Adding these all together, we see that

#(6z^2-4z+1)(8-3z)=color(green)(6z^2(8-3z))+color(blue)((-4z)(8-3z))+color(red)(1(8-3z)#

Distributing each, we obtain

#=color(green)(48z^2-18z^3)+color(blue)(-32z+12z^2)+color(red)(8-3z)#

Now, to simplify, sort this by degree (combine like terms):

#=-18z^3+underbrace(48z^2+12z^2) _ (48+12=60)underbrace(-32z-3z) _ (-32-3=-35)+8#

#=-18z^3+60z^2-35z+8#