Why must number "b" be a factor of 24 in case P(b)=O where P(x) is some quadrinomial ( of degree 3) with a final constant term equal to 24?

Question

Reference : https://www.stewartcalculus.com/data/default/upfiles/AlgebraReview.pdf, page 5.

In James Stewart's Review Of Algebra (a chapter of one of his Calculus Books I believe), I read this:

Let $P(x)= x^3 - 3x^2 - 10x + 24$, and $P(b)= 0$; then $b$ must be a factor of 24.

This is preliminary work to apply the factor theorem ( for polynomials of degree 3 or more).

Could you please explain on which ground Stewart asserts that " b must be a factor of 24".

Is there some other theorem involved here, of which I would have no knowledge?

Answer 1

By the rational root theorem, if $b$ is rational and $P(b)=b^3-3b^2-10b+24=0$,

then $b$ is a factor of $24$.

Answer 2

In my opinion rather than searching for theorems, it's better to prove it yourself.

$P(b) = b^3 - 3b^2 - 10b +24 =0$

Or, $b(3b + 10 - b^2) = 24$.

From this equation, it's clear that $b$ is a factor of $24$. Hence it's simple.