What is 12x^2+16x<x^4 ?

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12x^2+16x<x^4

12 x^{2} + 16 x < x^{4}

x < - 2 or - 2 < x < 0 or x > 4

Interval Notation

(- \infty, - 2) \cup (- 2, 0) \cup (4, \infty)

Solution steps

12 x^{2} + 16 x < x^{4}

Rewrite in standard form

12 x^{2} + 16 x - x^{4} < 0

Factor

12 x^{2} + 16 x - x^{4} : - x (x + 2)^{2} (x - 4)

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x < - 2 or - 2 < x < 0 or x > 4

4 (2 n - 4) + 3 = 8 n - 19

6 x + 8 = 2 x - 12

\frac{7}{9} + \frac{2}{3} - \frac{4}{5}

x^{4} + 2 x^{3} - 2 x - 1

(x^{2} + x + 1) (x^{2} + x + 1)