What is ((1-x^2)(2-x))/(x^2+5x+6)<= 0 ?

The solution to ((1-x^2)(2-x))/(x^2+5x+6)<= 0 is x<-3\lor-2<x<=-1\lor 1<= x<= 2

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((1-x^2)(2-x))/(x^2+5x+6)<= 0

\frac{( 1 - x ^{2} ) ( 2 - x )}{x ^{2} + 5 x + 6} \leq 0

x < - 3 or - 2 < x \leq - 1 or 1 \leq x \leq 2

Interval Notation

(- \infty, - 3) \cup (- 2, - 1] \cup [1, 2]

Solution steps

\frac{( 1 - x ^{2} ) ( 2 - x )}{x ^{2} + 5 x + 6} \leq 0

Factor

\frac{( 1 - x ^{2} ) ( 2 - x )}{x ^{2} + 5 x + 6} : \frac{( x + 1 ) ( x - 1 ) ( x - 2 )}{( x + 2 ) ( x + 3 )}

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

Identify the intervals

x < - 3 or - 2 < x \leq - 1 or 1 \leq x \leq 2

(x^{2} y)^{4}

x^{2} - x - 210 = 0

2 \leq x \leq 4

- 8 x + 3 \geq 27

\frac{3.8 - 2.54}{0.42}

What is ((1-x^2)(2-x))/(x^2+5x+6)<= 0 ?
The solution to ((1-x^2)(2-x))/(x^2+5x+6)<= 0 is x<-3\lor-2<x<=-1\lor 1<= x<= 2