What is 16x^3+16x^2>= x+1 ?

The solution to 16x^3+16x^2>= x+1 is -1 = 1/4

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16x^3+16x^2>= x+1

16 x^{3} + 16 x^{2} \geq x + 1

- 1 \leq x \leq - \frac{1}{4} or x \geq \frac{1}{4}

Interval Notation

[- 1, - \frac{1}{4}] \cup [\frac{1}{4}, \infty)

Decimal Notation

- 1 \leq x \leq - 0.25 or x \geq 0.25

Solution steps

16 x^{3} + 16 x^{2} \geq x + 1

Rewrite in standard form

16 x^{3} + 16 x^{2} - 1 - x \geq 0

Factor

16 x^{3} + 16 x^{2} - 1 - x : (x + 1) (4 x + 1) (4 x - 1)

(x + 1) (4 x + 1) (4 x - 1) \geq 0

Identify the intervals

- 1 \leq x \leq - \frac{1}{4} or x \geq \frac{1}{4}

x^{2} + 2 x - 57 = x - 1

3 (x - 5) = x + 11

(- 2 - 2)^{3}

35 + 30 x

2 n + 6 = 18

What is 16x^3+16x^2>= x+1 ?
The solution to 16x^3+16x^2>= x+1 is -1<= x<=-1/4 \lor x>= 1/4