What is 4x(1-x^2)(x^2-5x+6)>0 ?

The solution to 4x(1-x^2)(x^2-5x+6)>0 is x<-1\lor 0<x<1\lor 2<x<3

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4x(1-x^2)(x^2-5x+6)>0

4 x (1 - x^{2}) (x^{2} - 5 x + 6) > 0

x < - 1 or 0 < x < 1 or 2 < x < 3

Interval Notation

(- \infty, - 1) \cup (0, 1) \cup (2, 3)

Solution steps

4 x (1 - x^{2}) (x^{2} - 5 x + 6) > 0

Rewrite in standard form

- x^{5} + 5 x^{4} - 5 x^{3} - 5 x^{2} + 6 x > 0

Factor

- x^{5} + 5 x^{4} - 5 x^{3} - 5 x^{2} + 6 x : - x (x + 1) (x - 1) (x - 2) (x - 3)

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

Multiply both sides by

- 1

(reverse the inequality)

(- x (x + 1) (x - 1) (x - 2) (x - 3)) (- 1) < 0 \cdot (- 1)

Simplify

x (x + 1) (x - 1) (x - 2) (x - 3) < 0

Identify the intervals

x < - 1 or 0 < x < 1 or 2 < x < 3

3 (x + 2)^{2} - 2 (3 x - 1)^{2} - 3 x (x + 4) > - 38

\frac{( x + 1 )}{( x - 1 )} \geq \frac{( x + 5 )}{( x + 1 )}

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

(10 x - 7) (x - 5) \leq 0

- 3 x^{2} + 7 x + 6 > 0

What is 4x(1-x^2)(x^2-5x+6)>0 ?
The solution to 4x(1-x^2)(x^2-5x+6)>0 is x<-1\lor 0<x<1\lor 2<x<3