What is ((2x-x^2))/(((3+x)(5-x)))<0 ?

The solution to ((2x-x^2))/(((3+x)(5-x)))<0 is -3<x<0\lor 2<x<5

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((2x-x^2))/(((3+x)(5-x)))<0

\frac{( 2 x - x ^{2} )}{( ( 3 + x ) ( 5 - x ) )} < 0

- 3 < x < 0 or 2 < x < 5

Interval Notation

(- 3, 0) \cup (2, 5)

Solution steps

\frac{2 x - x ^{2}}{( 3 + x ) ( 5 - x )} < 0

Factor

\frac{2 x - x ^{2}}{( 3 + x ) ( 5 - x )} : \frac{- x ( x - 2 )}{( 3 + x ) ( 5 - x )}

\frac{- x ( x - 2 )}{( 3 + x ) ( 5 - x )} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - x ( x - 2 ) ) ( - 1 )}{( 3 + x ) ( 5 - x )} > 0 \cdot (- 1)

Simplify

\frac{x ( x - 2 )}{( 3 + x ) ( 5 - x )} > 0

Identify the intervals

- 3 < x < 0 or 2 < x < 5

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

\frac{( x - 1 ) ( x - 5 )}{( x - 3 )} > 0

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

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

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

What is ((2x-x^2))/(((3+x)(5-x)))<0 ?
The solution to ((2x-x^2))/(((3+x)(5-x)))<0 is -3<x<0\lor 2<x<5