What is x^3+7x^2>= 9x+63 ?

The solution to x^3+7x^2>= 9x+63 is -7 = 3

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x^3+7x^2>= 9x+63

x^{3} + 7 x^{2} \geq 9 x + 63

- 7 \leq x \leq - 3 or x \geq 3

Interval Notation

[- 7, - 3] \cup [3, \infty)

Solution steps

x^{3} + 7 x^{2} \geq 9 x + 63

Rewrite in standard form

x^{3} + 7 x^{2} - 63 - 9 x \geq 0

Factor

x^{3} + 7 x^{2} - 63 - 9 x : (x + 7) (x + 3) (x - 3)

(x + 7) (x + 3) (x - 3) \geq 0

Identify the intervals

- 7 \leq x \leq - 3 or x \geq 3

a^{4} + 14 a^{2} + 49

2 (x + 5) = 3 (x - 1) + 17

x^{2} + 2 x - 63

\frac{7}{9} - \frac{2}{9}

\frac{10 - 5}{5 + 2}

What is x^3+7x^2>= 9x+63 ?
The solution to x^3+7x^2>= 9x+63 is -7<= x<=-3\lor x>= 3