How do you multiply #(4z+3)(z-2)#?

1 Answer
Jun 27, 2016

#(4z+3)(z-2) = 4z^2-5z-6#

Explanation:

You have to use the laws of addition and multiplication:
Let me remind you some of these laws.
Commutative law of addition: #A+B=B+A#
Commutative law of multiplication: #A*B=B*A#
Associative law of addition: #(A+B)+C=A+(B+C)#
Associative law of multiplication: #(A * B) * C=A * (B * C)#
Distributive law: #(A+B)*C=A*C+B*C#

Using distributive law, consider
#A=4z#
#B=3#
#C=(z-2)#
Then
#(4z+3)(z-2) = 4z(z-2)+3(z-2)#

Use distributive law again for each term:
#4z(z-2) = 4z^2-8z#
#3(z-2)=3z-6#
Hence,
#(4z+3)(z-2) = 4z^2-8z+3z-6 = 4z^2-5z-6#

NOTE:
We could've started with distributive law by considering
#A=z#
#B=-2#
#C=4z+3#
Then
#(4z+3)(z-2) = (4z+3)z+(4z+3)(-2)#

Another application of distributive law:
#(4z+3)z = 4z^2+3z#
#(4z+3)(-2) = -8z-6#
Hence,
#(4z+3)(z-2) = 4z^2+3z-8z-6 = 4z^2-5z-6#

No surprise, the answer is exactly the same.