What is x^2-16>(x+4)(x^2-3x-1) ?

The solution to x^2-16>(x+4)(x^2-3x-1) is x<-4\lor 1<x<3

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x^2-16>(x+4)(x^2-3x-1)

x^{2} - 16 > (x + 4) (x^{2} - 3 x - 1)

x < - 4 or 1 < x < 3

Interval Notation

(- \infty, - 4) \cup (1, 3)

Solution steps

x^{2} - 16 > (x + 4) (x^{2} - 3 x - 1)

Rewrite in standard form

13 x - 12 - x^{3} > 0

Factor

13 x - 12 - x^{3} : - (x - 1) (x - 3) (x + 4)

- (x - 1) (x - 3) (x + 4) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (x - 3) (x + 4)) (- 1) < 0 \cdot (- 1)

Simplify

(x - 1) (x - 3) (x + 4) < 0

Identify the intervals

x < - 4 or 1 < x < 3

4 b (5 b - 3) - 2 (b^{2} - 7 b - 4)

n + 5 > - 3

216 x^{3} + 125 y^{3}

\frac{2 x - 1}{x + 1} = 3

2 x^{2} + 4 x + 3 \geq x^{2} - 2 x - 5

What is x^2-16>(x+4)(x^2-3x-1) ?
The solution to x^2-16>(x+4)(x^2-3x-1) is x<-4\lor 1<x<3