What is (x^2-4x-12)/(x-4)<= 0 ?

The solution to (x^2-4x-12)/(x-4)<= 0 is x<=-2\lor 4<x<= 6

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(x^2-4x-12)/(x-4)<= 0

\frac{x ^{2} - 4 x - 12}{x - 4} \leq 0

x \leq - 2 or 4 < x \leq 6

Interval Notation

(- \infty, - 2] \cup (4, 6]

Solution steps

\frac{x ^{2} - 4 x - 12}{x - 4} \leq 0

Factor

\frac{x ^{2} - 4 x - 12}{x - 4} : \frac{( x + 2 ) ( x - 6 )}{x - 4}

\frac{( x + 2 ) ( x - 6 )}{x - 4} \leq 0

Identify the intervals

x \leq - 2 or 4 < x \leq 6

4 (2 x + 7) + 5 (2 x + 7)

x^{2} - 2 x - 8 = 0

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

(x + 9)^{2} (x + 4) > 0

\frac{1 - 2 x}{x + 1} \geq 2 x - 1

What is (x^2-4x-12)/(x-4)<= 0 ?
The solution to (x^2-4x-12)/(x-4)<= 0 is x<=-2\lor 4<x<= 6