I just discovered the rational root theorem and I feel like I can understand it if I can get past the notational jargon presented in Wikipedia.
Here's what I think I understood. If $r$ is a rational solution of a polynomial $f(x)$ then the function can be rewritten as $(x-r)(g(x))$ where $g(x)$ is another polynomial. Expanding thus with $r$ places it in the constant term. Assuming for $g(x) \implies a, f(x) \implies -ar \quad \text{As the constant}$. So r is divisible by the constant term.
First of all, is that correct. And is there a way to explain the proof for the theorem without scary notation?
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$\begingroup$The statement of the rational root theorem is that if polynomial with integer coeficients is $P(x) =a_nx^n + a_{n-1}x^{n-1} + ...... + a_1x + a_0$ has a rational root (bear in mind it may have irrational roots) the ration root, call it $r$, will but be of the form $r = \frac jk$ where $j$ is a factor of $a_0$ and $k$ is a factor of $a_n$.
For instance if $2x^2 -9x + 4$ has any rational roots (it might not) then, as factors of $4$ are $\pm 1,\pm 2, \pm 4$ and the factors of $2$ are $\pm 1, \pm 2$, then the rational roots will be of the form $\frac jk$ where $j = \pm 1, \pm 2, \pm4$ and $k = \pm 1, \pm 2$ and, indeed, $r_1 = 4=\frac 41$ is one root, and $r_2 = \frac 12 =\frac 12$ is another.
And the proof can go like this:
Let $P(x) = a_nx^n + ..... + a_0$. Now for any $b$ we can do synthetic division to divide $P(x)$ but $x-b$ to get some $\frac {P(x)}{x-b} = c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+ .... + c_1x + c_0 + \frac s{x-b}$ and if we write $c_{n-1}x^{n-2} + c_{n-2}x^{n-2}+ .... + c_1x + c_0$ as $Q(x)$ we get $P(x) = (x-b)Q(x) + s$ where $s$ is a remainder constant.
Now if it turns out that $b$ was one of the roots of $P(x)$ we would have $P(b) = (b-b)Q(b) + s= s$. That means $s= 0$ and $P(x) = (x-b)Q(x)$.
ANd that means that $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ and $P(x) = (x- b)Q(x) = (x-b)(c_{n-1}x^{n-1} + c_{n-2}x^{n-2}+ .... + c_1x + c_0) = c_{n-1}x^n + (c_{n-2}-bc_{n-1})x^{n-1} + ....... +(c_0 -bc_1)x + c_0b$.
And that means $c_0b = a_0$ and so $b=\frac {a_0}{c_0}$. If $b$ is rational then $\frac {a_0}{c_0}$ when reduced to lowest terms will have a factor of $a_0$ in the numerator.
The also means $c_{n-1} = a_n$ and $c_{n-2}-bc_{n-1} = c_{n-2}-ba_n = a_{n-1}$.
This means $b = \frac {c_{n-2}-a_{n-1}}{a_n}$. And if $b$ is rational then $ \frac {c_{n-2}-a_{n-1}}{a_n}$ when reduced to lowest terms will have a factor of $a_n$ in the denominator.
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I've never done this before but let's see if we do an example:
Take $P(x) = 2x^2 -9x + 4$ and divide it by $x-\frac 12$.
We get $P(x) = 2x^2 - 9x + 4 = 2x(x-\frac 12) + x -9x +4 =$
$2x(x-\frac 12) - 8x + 4 = 2x(x-\frac 12) - 8(x-\frac 12) +(-4) + 4=$
$2x(x-\frac 12) - 8(x-\frac 12) = (x-\frac 12)(2x -8)$.
Now if we expand that out we will have $Ax^2 + Bx + C$ where:
$A = 2*1=2$ and $B=-8-\frac 12*2= -8-r*A$ and $C=-\frac 12*(-8)=-r*(-8)$.
So $\frac 12=r =\frac {-8-B}{A}$. so we must have the denominator of $r$ divide $A$. And, indeed, the denominator of $r$ is $2$ and $A$ is $2$.
And $\frac 12=r=\frac {-C}{-8}$ so we must have the numerator of $r$ divide $C$. And indeed, then numerator of $r$ is $1$ and $C= 4$.
$\endgroup$ 1 $\begingroup$Take your polynomial $p(x) = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n = 0$. We can take the $a_i$ as having no common factor, as we could divide it out. If $u/v$ is a rational zero of $p(x)$, written in lowest terms, substitute and multiply by $v^n$:
$\begin{align*} a_0 + a_1 \frac{u}{v} + a_2 \frac{u^2}{v^2} + \dotsb + a_n \frac{u^n}{v^n} &= 0 \\ a_0 v^n + a_1 u v^{n - 1} + a_2 u^2 v^{n - 2} + \dotsb + a_n u^n &= 0 \end{align*}$
All terms are integers. The second to the last terms are all divisible by $u$, so the first one must be divisible by $u$ also, thus $a_0$ is divisible by $u$. In the same way, considering the first up to the next to last terms and $v$, $a_n$ is divisible by $v$.
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