What is ((x^2-1))/(x^2+5x+4)<= 0 ?

The solution to ((x^2-1))/(x^2+5x+4)<= 0 is -4<x<-1\lor-1<x<= 1

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((x^2-1))/(x^2+5x+4)<= 0

\frac{( x ^{2} - 1 )}{x ^{2} + 5 x + 4} \leq 0

- 4 < x < - 1 or - 1 < x \leq 1

Interval Notation

(- 4, - 1) \cup (- 1, 1]

Solution steps

\frac{x ^{2} - 1}{x ^{2} + 5 x + 4} \leq 0

Factor

\frac{x ^{2} - 1}{x ^{2} + 5 x + 4} : \frac{( x + 1 ) ( x - 1 )}{( x + 1 ) ( x + 4 )}

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

Identify the intervals

- 4 < x < - 1 or - 1 < x \leq 1

9 x^{2} - 100 y^{2}

8 c^{2} + 7 c

4 u^{2} - 6 = 5 u

\frac{2}{3 - 1}

2 (- 9.8)

What is ((x^2-1))/(x^2+5x+4)<= 0 ?
The solution to ((x^2-1))/(x^2+5x+4)<= 0 is -4<x<-1\lor-1<x<= 1