What is ((x+5)(2x-7))/((x+1)^2)<= 0 ?

The solution to ((x+5)(2x-7))/((x+1)^2)<= 0 is -5<= x<-1\lor-1<x<= 7/2

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((x+5)(2x-7))/((x+1)^2)<= 0

\frac{( x + 5 ) ( 2 x - 7 )}{( x + 1 ) ^{2}} \leq 0

- 5 \leq x < - 1 or - 1 < x \leq \frac{7}{2}

Interval Notation

[- 5, - 1) \cup (- 1, \frac{7}{2}]

Decimal Notation

- 5 \leq x < - 1 or - 1 < x \leq 3.5

Solution steps

\frac{( x + 5 ) ( 2 x - 7 )}{( x + 1 ) ^{2}} \leq 0

Identify the intervals

- 5 \leq x < - 1 or - 1 < x \leq \frac{7}{2}

u^{2} - 10 u + 25

5 t + 30

- 4.8 (6.3 x - 4.18) = - 58.56

2 a + 5 \geq 17

5 y - 2 (4 - y) = 27 + 1 - 2 (2 y - 4) - 1

What is ((x+5)(2x-7))/((x+1)^2)<= 0 ?
The solution to ((x+5)(2x-7))/((x+1)^2)<= 0 is -5<= x<-1\lor-1<x<= 7/2