How do you multiply #(2m+3)(m-1)#?

2 Answers
Fir3hawk
May 20, 2017

#2m^2 +m -3#

Explanation:

First of all, we use the FOIL method, which helps when distributing factored binomials.
F: Multiply first terms of each parenthesis
O: Multiply the outside terms (first term of the first parenthesis and
last term of the second parenthesis)
I: Multiply the inside terms (last term of the first parenthesis and first term of the second parenthesis)
L: Multiply last terms of each parenthesis

Now, to the math:

#(2m+3)(m-1)#

F: #(2m) * (m) = 2m^2#

O: #(2m)*(-1) = -2m#

I: #(3)*(m) = 3m#

L: #(3)*(-1) = -3#

Now that we have all of our terms, we simplify to get the solution by adding like terms:

#(2m^2) + (-2m) + (3m) + (-3)#
= #2m^2 + (3m-2m) -3#
= #2m^2 +m -3#

Jim G.
May 20, 2017

#2m^2+m-3#

Explanation:

Each term in the second bracket is multiplied by each term in the first bracket as shown below.

#(color(red)(2m+3))(m-1)#

#=color(red)(2m)(m-1)color(red)(+3)(m-1)#

#=(color(red)(2m)xxm)+(color(red)(2m)xx-1)+(color(red)(3)xxm)+(color(red)(3)xx-1)#

#=2m^2+(-2m)+3m+(-3)#

#=2m^2-2m+3m-3#

#=2m^2+m-3#

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