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.