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 x<-3\lor-2<= x<-3/2 \lor-5/4 <= x<= 1/2

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(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} \leq 0

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

Interval Notation

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

Decimal Notation

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

Solution steps

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

Identify the intervals

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

\frac{( 3 \cdot 5 + 3 )}{10 - 3}

y^{2} + 7 y + 12

- 3 (3 x - 4) = 2 x + 23

(a^{2})^{2}

2 - (3 - 2 (x + 1)) = 3 x + 2 (x - (3 + 2 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 x<-3\lor-2<= x<-3/2 \lor-5/4 <= x<= 1/2