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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.

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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 $24$.

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  • $\begingroup$ 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 $\endgroup$ – Julian Drago Jun 21 '17 at 19:35
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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 $f$.

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  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jun 21 '17 at 6:35
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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 $24$.

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  • $\begingroup$ But it is possible that not all the roots are integers. $\endgroup$ – Eric Wofsey Jun 21 '17 at 6:34
  • $\begingroup$ @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. $\endgroup$ – Abhinav Dhawan Jun 21 '17 at 6:39
  • $\begingroup$ 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 $24$. $\endgroup$ – Eric Wofsey Jun 21 '17 at 6:39
  • $\begingroup$ Sorry... I got the point. Thanks for telling me... $\endgroup$ – Abhinav Dhawan Jun 21 '17 at 6:40

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