What is ((x^3*(x-2)(5-x)))/(((x^2-4)(x+1)))>0 ?

The solution to ((x^3*(x-2)(5-x)))/(((x^2-4)(x+1)))>0 is -2<x<-1\lor 0<x<2\lor 2<x<5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x^3*(x-2)(5-x)))/(((x^2-4)(x+1)))>0

\frac{( x ^{3} \cdot ( x - 2 ) ( 5 - x ) )}{( ( x ^{2} - 4 ) ( x + 1 ) )} > 0

- 2 < x < - 1 or 0 < x < 2 or 2 < x < 5

Interval Notation

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

Solution steps

\frac{x ^{3} ( x - 2 ) ( 5 - x )}{( x ^{2} - 4 ) ( x + 1 )} > 0

Factor

\frac{x ^{3} ( x - 2 ) ( 5 - x )}{( x ^{2} - 4 ) ( x + 1 )} : \frac{x ^{3} ( x - 2 ) ( 5 - x )}{( x + 2 ) ( x - 2 ) ( x + 1 )}

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

Identify the intervals

- 2 < x < - 1 or 0 < x < 2 or 2 < x < 5

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

\frac{1 + 12}{( x ^{2} )} < \frac{7}{x}

x^{2} - 4 x + 9 > 0

(x - 1) (x - 2) (x - 4) (x - 5) > 40

(x - 2) (x - 4) (x - 6) > 0

What is ((x^3*(x-2)(5-x)))/(((x^2-4)(x+1)))>0 ?
The solution to ((x^3*(x-2)(5-x)))/(((x^2-4)(x+1)))>0 is -2<x<-1\lor 0<x<2\lor 2<x<5