Asked by Linda
x^3+2x^2-x=2 Find the real number of solutions of the equation
Answers
Answered by
Dr Russ
So we need to factorise
x^3+2x^2-x-2=0
If we start off by guessing that one of the solutions is x=-1, i.e. one of the factors is (x+1)
then by division
(x^3+2x^2-x-2)/(x+1) = (x^2+x-2)
for which the factors are
(x-1)(x+2)
so the solutions are
x=-1, x=+1, x=-2
x^3+2x^2-x-2=0
If we start off by guessing that one of the solutions is x=-1, i.e. one of the factors is (x+1)
then by division
(x^3+2x^2-x-2)/(x+1) = (x^2+x-2)
for which the factors are
(x-1)(x+2)
so the solutions are
x=-1, x=+1, x=-2
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