(x^2+3x+1)(x^2+3x-3)>= 0

Lösung

(x^{2} + 3 x + 1) (x^{2} + 3 x - 3) \geq 0

x \leq \frac{- 3 - 21}{2} or \frac{- 3 - 5}{2} \leq x \leq \frac{5 - 3}{2} or x \geq \frac{21 - 3}{2}

Intervall-Notation

(- \infty, \frac{- 3 - 21}{2}] \cup [\frac{- 3 - 5}{2}, \frac{5 - 3}{2}] \cup [\frac{21 - 3}{2}, \infty)

Dezimale

x \leq - 3.79128 \dots or - 2.61803 \dots \leq x \leq - 0.38196 \dots or x \geq 0.79128 \dots

Schritte zur Lösung

(x^{2} + 3 x + 1) (x^{2} + 3 x - 3) \geq 0

Faktor

x^{2} + 3 x + 1 : (x - \frac{- 3 + 5}{2}) (x - \frac{- 3 - 5}{2})

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

Faktor

x^{2} + 3 x - 3 : (x - \frac{- 3 + 21}{2}) (x - \frac{- 3 - 21}{2})

(x - \frac{- 3 + 5}{2}) (x - \frac{- 3 - 5}{2}) (x - \frac{- 3 + 21}{2}) (x - \frac{- 3 - 21}{2}) \geq 0

Identifiziere die Intervalle

x \leq \frac{- 3 - 21}{2} or \frac{- 3 - 5}{2} \leq x \leq \frac{5 - 3}{2} or x \geq \frac{21 - 3}{2}

(x^{2} + 1) (x^{2} - 2 x + 5) \geq 0

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

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

\frac{( x ^{2} + 7 x + 10 )}{( x ^{2} - 4 )} \geq 0

6 x^{3} + x^{2} - 54 x + 56 < 0