(x^2+2x-8)(x+1)=0

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Solution for (x^2+2x-8)(x+1)=0 equation: 


Simplifying
(x2 + 2x + -8)(x + 1) = 0

Reorder the terms:
(-8 + 2x + x2)(x + 1) = 0

Reorder the terms:
(-8 + 2x + x2)(1 + x) = 0

Multiply (-8 + 2x + x2) * (1 + x)
(-8(1 + x) + 2x * (1 + x) + x2(1 + x)) = 0
((1 * -8 + x * -8) + 2x * (1 + x) + x2(1 + x)) = 0
((-8 + -8x) + 2x * (1 + x) + x2(1 + x)) = 0
(-8 + -8x + (1 * 2x + x * 2x) + x2(1 + x)) = 0
(-8 + -8x + (2x + 2x2) + x2(1 + x)) = 0
(-8 + -8x + 2x + 2x2 + (1 * x2 + x * x2)) = 0
(-8 + -8x + 2x + 2x2 + (1x2 + x3)) = 0

Combine like terms: -8x + 2x = -6x
(-8 + -6x + 2x2 + 1x2 + x3) = 0

Combine like terms: 2x2 + 1x2 = 3x2
(-8 + -6x + 3x2 + x3) = 0

Solving
-8 + -6x + 3x2 + x3 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

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