What is (x^2(3-x)(2x+1))/((x+4)(x+2))>= 0 ?

The solution to (x^2(3-x)(2x+1))/((x+4)(x+2))>= 0 is -4<x<-2\lor-1/2 <= x<= 3

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(x^2(3-x)(2x+1))/((x+4)(x+2))>= 0

\frac{x ^{2} ( 3 - x ) ( 2 x + 1 )}{( x + 4 ) ( x + 2 )} \geq 0

- 4 < x < - 2 or - \frac{1}{2} \leq x \leq 3

Interval Notation

(- 4, - 2) \cup [- \frac{1}{2}, 3]

Decimal Notation

- 4 < x < - 2 or - 0.5 \leq x \leq 3

Solution steps

\frac{x ^{2} ( 3 - x ) ( 2 x + 1 )}{( x + 4 ) ( x + 2 )} \geq 0

Identify the intervals

- 4 < x < - 2 or - \frac{1}{2} \leq x \leq 3

- 5 (x - 6) < 8 (- 4 x - 2)

5 x > - 5

- x^{2} + x + 2 > 0

x^{2} + x = 4

2 x^{2} = 64

What is (x^2(3-x)(2x+1))/((x+4)(x+2))>= 0 ?
The solution to (x^2(3-x)(2x+1))/((x+4)(x+2))>= 0 is -4<x<-2\lor-1/2 <= x<= 3