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

Interval Notation

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

Decimal Notation

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

k^{4} + k^{3} - k^{2}

\frac{3 ^{4}}{3 ^{- 2}}

x^{2} + 10 x = 21

x^{2} - 7 x + 25 = 0

ln (3 x) \leq 5

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