x^3+2x^2-x=2 Find the real number of solutions of the equation

Answer 1

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