Question

What are the real zeros of the function g(x) = x3 + 2x2 − x − 2?

Answer 1

For this case, to find the roots of the function, we equate to zero.

$x ^ 3 + 2x ^ 2-x-2 = 0$

We rewrite how:

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

We factor the maximum common denominator of each group:

$x ^ 2 (x + 2) -1 (x + 2) = 0$

We factor the polynomial, factoring the maximum common denominator$(x + 2):$

$(x + 2) (x ^ 2-1) = 0$

By definition of perfect squares we have to:

$a ^ 2-b ^ 2 = (a + b) (a-b)$

ON the expression$(x ^ 2-1):$

$a = x\\b = 1$

So:

$(x ^ 2-1) = (x + 1) (x-1)$

Thus, the factorization of the polynomial is:

$(x + 2) (x + 1) (x-1) = 0$

$x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1$

ANswer:$x_ {1} = - 2\\x_ {2} = - 1\\x_ {3} = 1$