What is ((x+1)(x-1)(x+2))/((x^2+3x+2)(x+5))<= 0 ?

The solution to ((x+1)(x-1)(x+2))/((x^2+3x+2)(x+5))<= 0 is -5<x<-2\lor-2<x<-1\lor-1<x<= 1

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((x+1)(x-1)(x+2))/((x^2+3x+2)(x+5))<= 0

\frac{( x + 1 ) ( x - 1 ) ( x + 2 )}{( x ^{2} + 3 x + 2 ) ( x + 5 )} \leq 0

- 5 < x < - 2 or - 2 < x < - 1 or - 1 < x \leq 1

Interval Notation

(- 5, - 2) \cup (- 2, - 1) \cup (- 1, 1]

Solution steps

\frac{( x + 1 ) ( x - 1 ) ( x + 2 )}{( x ^{2} + 3 x + 2 ) ( x + 5 )} \leq 0

Factor

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

\frac{( x + 1 ) ( x - 1 ) ( x + 2 )}{( x + 1 ) ( x + 2 ) ( x + 5 )} \leq 0

Identify the intervals

- 5 < x < - 2 or - 2 < x < - 1 or - 1 < x \leq 1

\frac{x ^{2}}{( x - 2 )} > 0

\frac{( 2 x + 1 )}{( 3 x - 6 )} \geq 3

100 - x^{2} \geq 0

\frac{( x - 1 )}{( x - 3 )} > 1

- x^{2} + 10 x - 21 \leq 0

What is ((x+1)(x-1)(x+2))/((x^2+3x+2)(x+5))<= 0 ?
The solution to ((x+1)(x-1)(x+2))/((x^2+3x+2)(x+5))<= 0 is -5<x<-2\lor-2<x<-1\lor-1<x<= 1