For this case, to find the roots of the function, we equate to zero.
[tex]x ^ 3 + 2x ^ 2-x-2 = 0[/tex]
We rewrite how:
[tex]x ^ 3 + 2x ^ 2- (x + 2) = 0[/tex]
We factor the maximum common denominator of each group:
[tex]x ^ 2 (x + 2) -1 (x + 2) = 0[/tex]
We factor the polynomial, factoring the maximum common denominator[tex](x + 2):[/tex]
[tex](x + 2) (x ^ 2-1) = 0[/tex]
By definition of perfect squares we have to:
[tex]a ^ 2-b ^ 2 = (a + b) (a-b)[/tex]
ON the expression[tex](x ^ 2-1):[/tex]
[tex]a = x\\b = 1[/tex]
So:
[tex](x ^ 2-1) = (x + 1) (x-1)[/tex]
Thus, the factorization of the polynomial is:
[tex](x + 2) (x + 1) (x-1) = 0[/tex]
[tex]x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1[/tex]
ANswer:[tex]x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1[/tex]
