What is (x+1)(x-5)(x^2+9)<= 0 ?

The solution to (x+1)(x-5)(x^2+9)<= 0 is -1<= x<= 5

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(x+1)(x-5)(x^2+9)<= 0

(x + 1) (x - 5) (x^{2} + 9) \leq 0

- 1 \leq x \leq 5

Interval Notation

[- 1, 5]

Solution steps

(x + 1) (x - 5) (x^{2} + 9) \leq 0

Identify the intervals

- 1 \leq x \leq 5

\frac{2 10}{3 11}

x^{2} - 6 x = 5

\frac{3}{10} + \frac{3}{10}

9 x^{2} - 18 x y + 5 y^{2}

(2 x + 1)^{7} + 4 x (x + 1) + 2

What is (x+1)(x-5)(x^2+9)<= 0 ?
The solution to (x+1)(x-5)(x^2+9)<= 0 is -1<= x<= 5