What is x^3+4x^2<= 4x+16 ?

The solution to x^3+4x^2<= 4x+16 is x<=-4\lor-2<= x<= 2

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x^3+4x^2<= 4x+16

x^{3} + 4 x^{2} \leq 4 x + 16

x \leq - 4 or - 2 \leq x \leq 2

Interval Notation

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

Solution steps

x^{3} + 4 x^{2} \leq 4 x + 16

Rewrite in standard form

x^{3} + 4 x^{2} - 16 - 4 x \leq 0

Factor

x^{3} + 4 x^{2} - 16 - 4 x : (x + 4) (x + 2) (x - 2)

(x + 4) (x + 2) (x - 2) \leq 0

Identify the intervals

x \leq - 4 or - 2 \leq x \leq 2

2 a^{2} - 13 a - 16 = 8

12 x^{2} - 11 x - 15

5 (z + 3.41) = - 7.89 (2 z - 4) - 5.67

m^{2} = 8 m - 15

\frac{- 2}{- 6}

What is x^3+4x^2<= 4x+16 ?
The solution to x^3+4x^2<= 4x+16 is x<=-4\lor-2<= x<= 2