What is (x^2+8x+12)/(x+4)>= 0 ?

The solution to (x^2+8x+12)/(x+4)>= 0 is -6 =-2

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(x^2+8x+12)/(x+4)>= 0

\frac{x ^{2} + 8 x + 12}{x + 4} \geq 0

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

Interval Notation

[- 6, - 4) \cup [- 2, \infty)

Solution steps

\frac{x ^{2} + 8 x + 12}{x + 4} \geq 0

Factor

\frac{x ^{2} + 8 x + 12}{x + 4} : \frac{( x + 2 ) ( x + 6 )}{x + 4}

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

Identify the intervals

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

2 x^{3} - 7 x^{2} - 4 x

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

\frac{1}{4} (5)^{- 1}

- 36 = 9 a - 63

- 202 + 20

What is (x^2+8x+12)/(x+4)>= 0 ?
The solution to (x^2+8x+12)/(x+4)>= 0 is -6<= x<-4\lor x>=-2