Why is it that: "If polynomials have integer coefficients, the roots of those polynomials will be a divisor (factor) of the constant term"?

Question

So, in my textbook, I came across a theorem/axiom (?) that states:

Given a polynomial $P(x)$ with integer coefficients, its roots, if they exist, are divisors of the independent/constant term of the polynomial

I've wrapped my head around this and I can't figure out why. There's a explanation in my book, but I can't quite understand it. It goes like this:

Given a polynomial $P(x)$ with whole numbers as coefficients

$P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}$ and $a$ is an integer root of $P(x)$ , then $a$ is a divisor of $a_{_{0}}$ (the independent term).

Demonstration:

$$P(a)=a_{n}a^n+a_{n-1}a^{n-1}\dots+a_{1}a+a_{0}=0$$

From here, it results that:

$$a_{0}=a_{n}a^n-a_{n-1}a^{n-1}-_{\dots}-a_{1}a$$ Therefore, $a_{0}$ is a multiple of $a$ , i. e. $a$ is a divisor of $a_{0}$ .

Can someone explain to me why this is true? And maybe give a real example with a polynomial.

Thanks.

Answer 1

The argument presented in the body of your question is perfectly sound [with a missing minus sign: a typo?] and is worth digesting. If you find that the notation is what's holding you back, then write out the entire argument in one special case but with different notation until you believe in it. This will make understanding the full result easier.

Let's argue this for quadratics only, so assume $P(x) = ax^2+bx+c$ where $a, b, c$ are all integers. Suppose $r$ is an integer root. Then $$ P(r) = ar^2 + br + c =0 $$ and so $$ c = -ar^2 - br = r(-ar -b). $$ It is now clear that $r$ is a divisor of $c$, the constant term.

Understanding this, it is not hard to extract the fully general argument for a polynomial of degree $n$. After all, the argument you've presented does the exact same thing, just with more terms and more generality.

Answer 2

Let $\mathit F(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$, where $\mathit a_n, a_{n-1}, \dots, a_0$ are all arbitrary integer coefficients.

$\it Note:$ $\;$The subscripts $\mathit n, n-1$, etc. don't mean that there's any relation between the coefficient and the power of each term, but rather just a convenient way of naming coefficients of polynomials of large degree instead of using all the letters of the alphabet or something along those lines.

Suppose r is an integral root of $\mathit F(x)$, $$\mathit \implies a_nr^n + a_{n-1}r^{n-1} + \dots + a_1r + a_0 = 0$$ $$\mathit \iff -a_nr^n - a_{n-1}r^{n-1} - \dots - a_1r = a_0$$ $$\mathit \iff r(-a_nr^{n-1} - a_{n-1}r^{n-2} - \dots - a_1) = a_0$$ $$\mathit \iff r\,|\, a_0$$

I'll give an example of a quadratic to clear up any confusion.

$$\mathit P(x) = x^2 + 8x + 7$$ $$\mathit P(x) = (x + 7)(x + 1)$$ $\mathit \implies x = -7$ and $\mathit x = -1$ are roots, which are factors of the constant term, i.e. 7.

Hopefully that helped clear up any confusion you had! You may want to check out something called the Rational Roots Theorem, which is essentially a more generalised case of this theorem that extends across the rational numbers instead of just the integers.