Proof of Theorem 43 (Roots of polynomials)

We will need the following theorem which allows us to divide polynomials. This is an analogue of the integer division theorem that we saw in the number theory chapter.

Theorem 45 (Euclidian division of polynomials)

Given two polynomials $f(x)$ and $g(x)\neq 0$, there exist unique polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that

$f(x) = q(x) g(x) + r(x)$

and $\deg(r) < \deg(g)$.

We will not prove this theorem.

Example 43

Notice that, when $\deg(g)$ is one (e.g., dividing by $g(x) = x-a$), the division theorem implies that $f(x) = q(x) (x-a) + r(x)$ where $r(x)$ is a constant (degree zero). We can deduce the constant by plugging in $f(a) = 0 + r(a)$, so the constant is $f(a)$. This implies that,

$f(x) = q(x) (x-a) + f(a)$

Below, we show an example of dividing $x^3+ 2x +1$ by $(x+1)$ (which should look intuitive).

$$\begin{array}{r|l} x+1 & x^3 + 0x^2 + 2x + 1 \\ \cdot x^2 & - (x^3 + x^2) \\ & \underline{0x^3 - x^2 + 2x + 1} \\ \cdot -x & \phantom{000}- (-x^2 - x) \\ & \underline{0x^2 + 3x + 1} \\ \cdot 3 & \phantom{000}- (3x + 3) \\ & \underline{0x + (-2)} \end{array}$$

So, the quotient is $x^2 - x + 3$ and the remainder is $-2$.

Lemma 15

If $a$ is a root of $f$, then there exists a unique polynomial $q$ such that $f(x) = (x-a) q(x)$.

Proof:

Using Example 43, if we divide $f(x)$ by $(x-a)$, we have that $f(x) = (x-a) q(x) + f(a)$. Since $a$ is a root, we have $f(a) = 0$.

Exercise 121

This exercise asks you to complete the proof of Theorem 43.

• Prove the following by induction on $d \geq 1$: Let $f$ be a polynomial of degree $d$ with distinct roots $r_1,\ldots, r_d$. Then $f(x) = c\cdot \prod_{i \in [d]} (x-r_i)$ for some constant $c$.
• Deduce Theorem 43 from the first part.