Understanding use of factor theorem to factor polynomial based on possible factors from constant

Question

I'm reviewing my algebra in prep for Linear Algebra, using the review of Algebra in Stewart's Single Variable Calculus, here: http://stewartcalculus.com/data/CALCULUS_8E_ET/upfiles/6et_reviewofalgebra.pdf

On p.4, they use the Factor Theorem to factor $$x^3 - 3x^2 -10x + 24$$

They state the following:

If P(b) = 0, where b is an integer, then b is a factor of 24.

They then use this fact to test possible integer values of b as a factor of 24 (+1, -1, +2, -2, etc.). What I'm not understanding is how they arrive at the fact that b must be a factor of 24 based on the the Factor Theorem. I understand from the Factor Theorem that if P(b) = 0, then x - b will be a factor of P(x), but how did they discern that b would be a factor of 24? Thanks in advance.

Answer 1

Because, by the rational root theorem, every rational root $q_0$ of a polynomial $a_0+a_1x+\cdots+a_nx^n$ with integer coefficients can be written as $\frac ab$, where $a$ and $b$ are integers such that $b$ divides $a_n$ and $a$ divides $a_0$. In your case, $a_n=1$ and $a_0=24$. So, $q$ must be an integer that divides $24$.

Answer 2

If $(x-b)$ is a factor, then $f(x)=g(x)(x-b)$, Assuming the constant term of $f$ is nonzero, then one must have that $b$ times the constant term of $g$ is the constant term of $f$. This tells us that $b$ must divide the constant term of $f$.

Answer 3

The product of all roots of this polynomial is $24$ as you can see. Also $b$ is a root of this polynomial as given $P(b)=0$. Therefore $b$ must be a factor of $24$.