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

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Answer 1 · 2018-06-02T21:52:49

#x=-1#

#x=-3#

#x=1#

#(x^2+2x)^2-2(x^2+2x)-3=0#

Let #k = x^2+2x#

Substute:

#k^2-2k-3=0#

Factor:

#(k+1)(k-3)= 0

Now un-substute:

#(x^2+2x+1)(x^2+2x-3)= 0

Factor:

#(x+1)(x+1)(x+3)(x-1)=0#

#(x+1)^2(x+3)(x-1)=0#

#x=-1#

#x=-3#

#x=1#

graph{(x+1)^2(x+3)(x-1) [-10, 10, -5, 5]}