What is -3x^3-12x^2-3x+18>= 0 ?

The solution to -3x^3-12x^2-3x+18>= 0 is x<=-3\lor-2<= x<= 1

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-3x^3-12x^2-3x+18>= 0

- 3 x^{3} - 12 x^{2} - 3 x + 18 \geq 0

x \leq - 3 or - 2 \leq x \leq 1

Interval Notation

(- \infty, - 3] \cup [- 2, 1]

Solution steps

- 3 x^{3} - 12 x^{2} - 3 x + 18 \geq 0

Rewrite in standard form

- x^{3} - 4 x^{2} - x + 6 \geq 0

Factor

- x^{3} - 4 x^{2} - x + 6 : - (x - 1) (x + 2) (x + 3)

- (x - 1) (x + 2) (x + 3) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (x + 2) (x + 3)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 1) (x + 2) (x + 3) \leq 0

Identify the intervals

x \leq - 3 or - 2 \leq x \leq 1

\frac{2}{\frac{6}{7}}

(\frac{3}{5})^{2}

14 s^{4} - 19 s^{2} r^{2} + 5 r^{4}

20 e^{5} f^{2} - 16 e^{2} f

(1 - i)^{20}

What is -3x^3-12x^2-3x+18>= 0 ?
The solution to -3x^3-12x^2-3x+18>= 0 is x<=-3\lor-2<= x<= 1