What is (x^3+3x^2-9x-27)/(x+4)<= 0 ?

The solution to (x^3+3x^2-9x-27)/(x+4)<= 0 is -4<x<= 3

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(x^3+3x^2-9x-27)/(x+4)<= 0

\frac{x ^{3} + 3 x ^{2} - 9 x - 27}{x + 4} \leq 0

- 4 < x \leq 3

Interval Notation

(- 4, 3]

Solution steps

\frac{x ^{3} + 3 x ^{2} - 9 x - 27}{x + 4} \leq 0

Factor

\frac{x ^{3} + 3 x ^{2} - 9 x - 27}{x + 4} : \frac{( x + 3 ) ^{2} ( x - 3 )}{x + 4}

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

Identify the intervals

- 4 < x \leq 3

x^{2} = 2 x + 10

3 y - 5 y - 18 = - 2

ln (\frac{1}{e ^{2}})

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

(1 + \frac{1}{3})^{2} - \frac{2}{9}

What is (x^3+3x^2-9x-27)/(x+4)<= 0 ?
The solution to (x^3+3x^2-9x-27)/(x+4)<= 0 is -4<x<= 3