What is (x^2+2x-3)(3x-4-x^2)>0 ?

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

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(x^2+2x-3)(3x-4-x^2)>0

(x^{2} + 2 x - 3) (3 x - 4 - x^{2}) > 0

- 3 < x < 1

Interval Notation

(- 3, 1)

Solution steps

(x^{2} + 2 x - 3) (3 x - 4 - x^{2}) > 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (x + 3) (x^{2} - 3 x + 4)) (- 1) < 0 \cdot (- 1)

Simplify

(x - 1) (x + 3) (x^{2} - 3 x + 4) < 0

Identify the intervals

- 3 < x < 1

(x^{2} - x - 1) (x^{2} - x - 7) < - 5

\frac{( 10 x )}{( x ^{2} + 9 )} \leq 1

\frac{e}{x} > 0

\frac{( 3 x ^{2} - 4 x - 6 )}{( 2 x - 5 )} < 0

\frac{( x ^{2} - 7 x + 12 )}{( 2 x ^{2} + 4 x + 5 )} > \frac{1}{2}

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