What is ((3x-12))/((x^2-2x-3))<= 0 ?

The solution to ((3x-12))/((x^2-2x-3))<= 0 is x<-1\lor 3<x<= 4

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((3x-12))/((x^2-2x-3))<= 0

\frac{( 3 x - 12 )}{( x ^{2} - 2 x - 3 )} \leq 0

x < - 1 or 3 < x \leq 4

Interval Notation

(- \infty, - 1) \cup (3, 4]

Solution steps

\frac{3 x - 12}{x ^{2} - 2 x - 3} \leq 0

Factor

\frac{3 x - 12}{x ^{2} - 2 x - 3} : \frac{3 ( x - 4 )}{( x + 1 ) ( x - 3 )}

\frac{3 ( x - 4 )}{( x + 1 ) ( x - 3 )} \leq 0

Divide both sides by

3

\frac{\frac{3 ( x - 4 )}{( x + 1 ) ( x - 3 )}}{3} \leq \frac{0}{3}

Simplify

\frac{x - 4}{( x + 1 ) ( x - 3 )} \leq 0

Identify the intervals

x < - 1 or 3 < x \leq 4

\frac{256}{( 2 ^{16} - 1 )} \cdot 100

\frac{1}{5} + \frac{2}{7} - \frac{6}{35}

- 8 x^{2} y^{3} (- 7 x^{2} y^{2})

b^{2} - 4 b + 4 = 0

(3)^{- 1}

What is ((3x-12))/((x^2-2x-3))<= 0 ?
The solution to ((3x-12))/((x^2-2x-3))<= 0 is x<-1\lor 3<x<= 4