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

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.

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 $24$.

• Thank you. Now I understand. Not surprisingly I also had forgotten the rational root theorem, so here's a video for anyone who stumbles on to this and is in the same predicament. youtube.com/watch?v=gs0S9LpuxmE – Julian Drago Jun 21 '17 at 19:35

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 $f$.

• There is no need to assume the constant term of $f$ is nonzero. On the other hand, it is necessary to know that $g(x)$ has integer coefficients (or at least integer constant term), which you can prove using the division algorithm. – Eric Wofsey Jun 21 '17 at 6:35

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 $24$.

• But it is possible that not all the roots are integers. – Eric Wofsey Jun 21 '17 at 6:34
• @Eric Wofsey But it is mentioned that b is an integer in the question asked. That's why I didn't mentioned that explicitly in my answer. – Abhinav Dhawan Jun 21 '17 at 6:39
• You know that $b$ times the other two roots is $24$. But if the other two roots are not integers, this does not tell you $b$ is a factor of $24$. – Eric Wofsey Jun 21 '17 at 6:39
• Sorry... I got the point. Thanks for telling me... – Abhinav Dhawan Jun 21 '17 at 6:40