How do you solve #8m ^ { 2} + 20m = 12# by factoring?

1 Answer
Jun 7, 2017

#m = 1/2 and m = -3#

Explanation:

Given: #8m^2 + 20m = 12#

First put the equation in #Ax^2 + Bx + C = 0# form.

#8m^2 + 20m - 12 = 0#

Next factor the greatest common factor (GCF) from each term:

#4(2m^2 + 5m - 3) = 0#

The factored equation will look like this: #4(2m + x)(m + y) = 0#

Now find two numbers #x, y# that multiply to #-3#:

#-3 and 1 " or " -1 and 3#.

Middle term using distribution will be: # x + 2y = 5#.

Let #x = 1, y = -3: " "2(-3) + 1 = -5" "# Doesn't work
Let #x = -3, y = 1: " "2(1) + (-3) = -1 " "# Doesn't work
Let #x = 3, y = -1: " "2(-1) + 3 = 1 " "# Doesn't work
Let #x = -1, y = 3: " "2(3) + (-1) = 5 " " # WORKS!

#4 (2m -1)(m + 3) = 0#

The two factors that make the equation #= 0# are:

#2m - 1 = 0 " and " m + 3 = 0#

#2m = 1 " and " m = -3#

#m = 1/2 " and " m = -3#