What is (x^2-3x-2)/(x^2+5x+6)<(2-x)/(x^2-4) ?

The solution to (x^2-3x-2)/(x^2+5x+6)<(2-x)/(x^2-4) is -3<x<-2

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(x^2-3x-2)/(x^2+5x+6)<(2-x)/(x^2-4)

\frac{x ^{2} - 3 x - 2}{x ^{2} + 5 x + 6} < \frac{2 - x}{x ^{2} - 4}

- 3 < x < - 2

Interval Notation

(- 3, - 2)

Solution steps

\frac{x ^{2} - 3 x - 2}{x ^{2} + 5 x + 6} < \frac{2 - x}{x ^{2} - 4}

Rewrite in standard form

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

Factor

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

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

Identify the intervals

- 3 < x < - 2

1 = 2 x^{2} + 7 x

- 4 x^{2} + 6 x + 20 = 0

\frac{1}{2} x^{2} - 5 x = 0

a x + x y + y^{2} + a y

x^{2} + 20 = 4

What is (x^2-3x-2)/(x^2+5x+6)<(2-x)/(x^2-4) ?
The solution to (x^2-3x-2)/(x^2+5x+6)<(2-x)/(x^2-4) is -3<x<-2