the problem is:
$\displaystyle x^3 + x^2 - 32x - 60$
It says something about guessing and checking, but how do I know if the number I pick works?
you can try using the rational roots theorem to check for rational zeros. the roots here might not be rational since they ask for bounds rather than the actual zeros, but who knows. start by plugging in the factors (+/-) of 60 to see if you get zero when you evaluate it. if you do, great, if not, try picking numbers close to that that get you as close to zero as possible from both sides, that will get you the bounds. if you actually find a rational zero, call it a, then you can divide the polynomial by (x - a) and it will reduce it to a quadratic where you can find the other zeros.
Because this is monic the rational root theorem tells us that if this has a rational root it is in fact an integer and a factor of $\displaystyle 24$.
The factors of 24 are $\displaystyle \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24$ so you try $\displaystyle x=\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24$ in turn untill you find one that works.
Once you have found a root r, you know that $\displaystyle (x-r)$ is a factor, and so you can divide the cubic through by the factor to find quadratic $\displaystyle x^2+ax+b$ such that:
$\displaystyle x^3 - 5x^2 - 2x + 24=(x-r)(x^2+ax+b)$
and carry on from there
RonL
i don't think i understand your question. all zeros will give a remainder of zero when you plug them in, or divide by the factor we pointed out. if you find a zero, you can reduce it to a quadratic in which case you can find the other zeros by factoring or using the quadratic formula. otherwise, try the guess and check approach i described