How do you solve $x^3-13x-12<=0$ ?

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Answer 1 · 2016-11-24T09:46:30

$x <=-3 and x in [-1, 4]$

Let $f(x) = x^3-13x-12$ . The sum of the coefficients is 0. Xo, x+1 is a

factor.

Let the other factor be $x^2+ax+b$ .

By comparing

the product with f(x), $a =-1 and b =-12$ .

The factors of $x^2-x-12 are x+3 and x-4$ . So,

$f(x) = (x+1)(x+3)(x-4) <=0$ .

One factor only or all the factors should be $<=0$ .

For $x <= -3 and x in [-1, 4]$ , this happens.

How do you solve x^3-13x-12<=0x^3-13x-12<=0?