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

The solution to ((2x-1))/((2x^3+3x^2+x))>0 is x 1/2

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

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

x < - 1 or - \frac{1}{2} < x < 0 or x > \frac{1}{2}

Interval Notation

(- \infty, - 1) \cup (- \frac{1}{2}, 0) \cup (\frac{1}{2}, \infty)

Decimal Notation

x < - 1 or - 0.5 < x < 0 or x > 0.5

Solution steps

\frac{2 x - 1}{2 x ^{3} + 3 x ^{2} + x} > 0

Factor

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

\frac{2 x - 1}{x ( 2 x + 1 ) ( x + 1 )} > 0

Identify the intervals

x < - 1 or - \frac{1}{2} < x < 0 or x > \frac{1}{2}

(x + 2) \cdot (3 x^{2} - 4 x - 7) < 0

x^{2} - 7 x^{2} + 14 x - 8 < 0

x - 1 < \frac{1}{( x - 3 )}

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

x^{2} > - 4

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