What is (x-2-x^2)(x^2+2x-8)<0 ?

The solution to (x-2-x^2)(x^2+2x-8) 2

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(x-2-x^2)(x^2+2x-8)<0

(x - 2 - x^{2}) (x^{2} + 2 x - 8) < 0

x < - 4 or x > 2

Interval Notation

(- \infty, - 4) \cup (2, \infty)

Solution steps

(x - 2 - x^{2}) (x^{2} + 2 x - 8) < 0

Factor

(x - 2 - x^{2}) (x^{2} + 2 x - 8) : - (x - 2) (x + 4) (x^{2} - x + 2)

- (x - 2) (x + 4) (x^{2} - x + 2) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 2) (x + 4) (x^{2} - x + 2)) (- 1) > 0 \cdot (- 1)

Simplify

(x - 2) (x + 4) (x^{2} - x + 2) > 0

Identify the intervals

x < - 4 or x > 2

0^{- \frac{1}{2}}

- \frac{3}{4} - 1

\frac{1350}{600}

- \frac{39}{56} - \frac{22}{35}

x^{3} + 5 x^{2} - 9 x - 45

What is (x-2-x^2)(x^2+2x-8)<0 ?
The solution to (x-2-x^2)(x^2+2x-8)<0 is x<-4\lor x>2