What is (8x^3+22x^2+7x-10)/(2x^2+9x+9)>= 0 ?

The solution to (8x^3+22x^2+7x-10)/(2x^2+9x+9)>= 0 is -3 = 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(8x^3+22x^2+7x-10)/(2x^2+9x+9)>= 0

\frac{8 x ^{3} + 22 x ^{2} + 7 x - 10}{2 x ^{2} + 9 x + 9} \geq 0

- 3 < x \leq - 2 or - \frac{3}{2} < x \leq - \frac{5}{4} or x \geq \frac{1}{2}

Interval Notation

(- 3, - 2] \cup (- \frac{3}{2}, - \frac{5}{4}] \cup [\frac{1}{2}, \infty)

Decimal Notation

- 3 < x \leq - 2 or - 1.5 < x \leq - 1.25 or x \geq 0.5

Solution steps

\frac{8 x ^{3} + 22 x ^{2} + 7 x - 10}{2 x ^{2} + 9 x + 9} \geq 0

Factor

\frac{8 x ^{3} + 22 x ^{2} + 7 x - 10}{2 x ^{2} + 9 x + 9} : \frac{( x + 2 ) ( 2 x - 1 ) ( 4 x + 5 )}{( 2 x + 3 ) ( x + 3 )}

\frac{( x + 2 ) ( 2 x - 1 ) ( 4 x + 5 )}{( 2 x + 3 ) ( x + 3 )} \geq 0

Identify the intervals

- 3 < x \leq - 2 or - \frac{3}{2} < x \leq - \frac{5}{4} or x \geq \frac{1}{2}

7 y - 9 = 24 - 4 y

- 12 x^{2} - 25 x - 12 = 0

7.0795 t = 57.34 t - 16.1 t^{2}

\frac{x}{2} \geq \frac{9}{x + 1} + 8

3 x + 4 \leq 6 - x

What is (8x^3+22x^2+7x-10)/(2x^2+9x+9)>= 0 ?
The solution to (8x^3+22x^2+7x-10)/(2x^2+9x+9)>= 0 is -3<x<=-2\lor-3/2 <x<=-5/4 \lor x>= 1/2