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
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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.
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