What is 15x+18<= x^2(2x+1) ?

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15x+18<= x^2(2x+1)

15 x + 18 \leq x^{2} (2 x + 1)

- 2 \leq x \leq - \frac{3}{2} or x \geq 3

Interval Notation

[- 2, - \frac{3}{2}] \cup [3, \infty)

Decimal Notation

- 2 \leq x \leq - 1.5 or x \geq 3

Solution steps

15 x + 18 \leq x^{2} (2 x + 1)

Rewrite in standard form

15 x + 18 - x^{2} - 2 x^{3} \leq 0

Factor

15 x + 18 - x^{2} - 2 x^{3} : - (x + 2) (2 x + 3) (x - 3)

- (x + 2) (2 x + 3) (x - 3) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 2) (2 x + 3) (x - 3)) (- 1) \geq 0 \cdot (- 1)

Simplify

(x + 2) (2 x + 3) (x - 3) \geq 0

Identify the intervals

- 2 \leq x \leq - \frac{3}{2} or x \geq 3

4 x^{2} + 4 x = 15

2 x^{2} - 9 x = - 5

a (12) = - 5 + 6 (12 - 1)

2 x^{2} - 6 = 0

- 4 - 5 v < - 29

What is 15x+18<= x^2(2x+1) ?
The solution to 15x+18<= x^2(2x+1) is -2<= x<=-3/2 \lor x>= 3