How do you solve #(x+1)^2(x+2)=0#?

1 Answer
Teun
Dec 16, 2016

There are two solutions: #x_1 = -2 " and "x_2 = -1#.

Explanation:

#(x+1)^2(x+2)=0#
Therefore,
#(x+1)^2 = 0 " or " x+2=0#

So our first value of x is:
#x_1+2=0 -> x_1=-2#

The two other values can be solved for by rewriting.
#(x+1)^2 = 0-> (x+1)(x+1) =0#
Because this is a perfect square there is only one solution.
So, #x_2+1 = 0 -> x_2 = -1#.

Our solutions are, #x_1 = -2 " and "x_2 = -1#.

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