How do you solve #(x+4)^3(x+2)^2=0#?

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Answer 1 · 2017-06-14T13:48:16

#x = - 4, - 2#

We have: #(x + 4)^(3) (x + 2)^(2) = 0#

Using the null factor law:

#Rightarrow (x + 4)^(3) = 0#

#Rightarrow x + 4 = 0#

#therefore x = - 4#

#or#

#Rightarrow (x + 2)^(2) = 0#

#Rightarrow x + 2 = 0#

#therefore x = - 2#

Therefore, the solutions to the equation are #x = - 4# and #x = - 2#.