# x^3+2x^2-x-2=0 solve for x

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

Substitute with 1:

1+2-1-2=0

Then 1 is one of the function's roots.

Then (x-1) is one of the factors.

Now divide the function by (x-1).

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

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

Factorize:

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

Then the roots are:

x1=1

x2=-1

x3=-2

Given:

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

We factorise the left hand side of the equation as follows:

x^3 + 2x^2 - x - 2

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

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

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

Therefore:

x = -1, 1, and -2

To solve x^3+2x^2-x-2 = 0

Solution:

We shall factorise the left side by grouping:

x^2+2x^2-x-2 = 0.

x^2(x+2) - 1(x+2) = 2. Pull out x+2 on the left:

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

(x+2)(x+1)(x-1) = 0. As x^2-1 = x^2-1^2 = (x+1)(x-1).

So x+2 = 0 or (x+1) = 0 Or (x-1) = 0. Or

x=-2, or x=-1, or x=-1 are the solutions.