What is ((2x+1)(3-x))/((x^2+6x+8)(x^2+1))<= 0 ?

The solution to ((2x+1)(3-x))/((x^2+6x+8)(x^2+1)) = 3

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((2x+1)(3-x))/((x^2+6x+8)(x^2+1))<= 0

\frac{( 2 x + 1 ) ( 3 - x )}{( x ^{2} + 6 x + 8 ) ( x ^{2} + 1 )} \leq 0

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

Solution steps

\frac{( 2 x + 1 ) ( 3 - x )}{( x ^{2} + 6 x + 8 ) ( x ^{2} + 1 )} \leq 0

Factor

\frac{( 2 x + 1 ) ( 3 - x )}{( x ^{2} + 6 x + 8 ) ( x ^{2} + 1 )} : \frac{( 2 x + 1 ) ( 3 - x )}{( x + 2 ) ( x + 4 ) ( x ^{2} + 1 )}

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

Identify the intervals

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

9 x^{4} - x = 40

\frac{( - 2 x y ) ^{3} ( 3 a b ^{2} ) ^{2} ( - a x ^{2} y ^{3} )}{( - 3 a b ^{2} ) ^{2} ( 2 x y ^{2} ) ^{3}}

(159) (0.64) (T - 277) = (260) (341 - T)

(3 k - 1)^{2} = - 8

4 r^{2} + 20 r + 25 = 0

What is ((2x+1)(3-x))/((x^2+6x+8)(x^2+1))<= 0 ?
The solution to ((2x+1)(3-x))/((x^2+6x+8)(x^2+1))<= 0 is x<-4\lor-2<x<=-1/2 \lor x>= 3