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

Lösung

\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}

Intervall-Notation

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

Dezimale

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

Schritte zur Lösung

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

Faktorisiere

\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

Identifiziere die Intervalle

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}

f a c t or y^{2} + 7 y + 12

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

s im pl i f y (a^{2})^{2}

2 - (3 - 2 (x + 1)) = 3 x + 2 (x - (3 + 2 x))