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

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

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((x^2+3x+2))/((x-4))<0

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

x < - 2 or - 1 < x < 4

Interval Notation

(- \infty, - 2) \cup (- 1, 4)

Solution steps

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

Factor

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

\frac{( x + 1 ) ( x + 2 )}{x - 4} < 0

Identify the intervals

x < - 2 or - 1 < x < 4

- 2 x^{2} + 15 x - 7

\frac{4}{7} \cdot 4

- (2 x + 1)^{6}

\frac{3 x ^{8} y ^{3}}{9 x ^{6} y ^{8}}

3 x - 4 = 5 x - 2

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