Properties of Polynomials Examples 2

Recall from the Properties of Polynomials page that a function of the form $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is called a polynomial, and if $a_m \neq 0$ then we say that the degree of the polynomial $p$ is $n$ written $\mathrm{deg} p = n$.

We said that a number $\lambda$ is a root of the polynomial $p$ if $p(\lambda) = 0$.

We then saw that $\lambda \in \mathbb{F}$ is a root of the polynomial $p$ with $\mathrm{deg} p = n ≥ 1$ if and only if there exists a polynomial $q$ where $\mathrm{deg} q = n - 1$ such that $p$ can be factored as:

(1)

\begin{align} \quad p(x) = (x - \lambda) q(x) \end{align}

From this, we saw that a polynomial $p$ with $\mathrm{deg} p = n$ can have at most $n$ distinct roots.

Lastly, we noted that a polynomial that is the zero function has its coefficients $a_0 = a_1 = ... = a_n = 0$.

We will now look at some more examples regarding polynomials.

Example 1

Show that the subset $\{ 0 \} \cup \{ p(x) \in \wp ( \mathbb{F} ) : \mathrm{deg} p \: \mathrm{is \: odd} \}$ of $\wp ( \mathbb{F} )$ is not a subspace of $\wp (\mathbb{F})$.

Consider the polynomials $p(x) = x^3 + x^2$ and $q(x) = -x^3$. Both of these polynomials have odd degree and therefore are contained in the subset $\{ 0 \} \cup \{ p(x) \in \wp ( \mathbb{F} ) : \mathrm{deg} p \: \mathrm{is \: odd} \}$.

Note that $p(x) = q(x) = x^3 + x^2 - x^3 = x^2$. $\mathrm{deg} (p + q) = 2$, and so $(p(x) + q(x)) \not \in \{ 0 \} \cup \{ p(x) \in \wp ( \mathbb{F} ) : \mathrm{deg} p \: \mathrm{is \: odd} \}$. Therefore $\{ 0 \} \cup \{ p(x) \in \wp ( \mathbb{F} ) : \mathrm{deg} p \: \mathrm{is \: odd} \}$ is not closed under addition and therefore not a subspace of $\wp (\mathbb{F})$.

Example 2

Let $\mathrm{deg} p = m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and $p'$ have no roots in common.

$\Rightarrow$ Suppose that $p$ has $m$ distinct roots, $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{F}$. Then for $c \in \mathbb{F}$ we can factor $p(x)$ as:

(2)

\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) \end{align}

Take a root $\lambda_j$ for $j = 1, 2, ..., m$. In specific, we can factor $p(x)$ as:

(3)

\begin{align} \quad p(x) = (x - \lambda_j)q(x) \end{align}

The polynomial $q(x) = (x - \lambda_1)(x - \lambda_2)...(x - \lambda_{j-1})(x - \lambda_{j+1})...(x - \lambda_m)$. Now if we differentiate both sides of the equation above, we have that:

(4)

\begin{align} \quad p'(x) = (x - \lambda_j)q'(x) + q(x) \end{align}

Thus we have that:

(5)

\begin{align} \quad p'(\lambda_j) = (\lambda_j - \lambda_j)q'(\lambda_j) + q(\lambda_j) \\ \quad p'(\lambda_j) = q(\lambda_j) \neq 0 \end{align}

We can apply the same logic to the other roots of $p$ and we see that $\lambda_1, \lambda_2, ..., \lambda_m$ are not roots of $p'$ so $p$ and $p'$ have no roots in common.

$\Leftarrow$ Suppose that $p$ and $p'$ have no roots in common. Proving this can be simplified by proving the logically equivalent contrapositive which says that if $p$ does not have $m$ distinct roots then $p$ and $p'$ have a root in common.

Suppose that $p$ does not have $m$ distinct roots. Then some root $\lambda \in \mathbb{F}$ must have multiplicity greater than $1$, say $\lambda$ has multiplicity $k$. Then for some polynomial $q(x)$ we can factor $p(x)$ as:

(6)

\begin{align} \quad p(x) = (x - \lambda)^k q(x) \end{align}

If we differentiate both sides of the equation above we have that:

(7)

\begin{align} \quad p'(x) = k(x - \lambda)^{k-1}q(x) + (x - \lambda)^kq'(x) \end{align}

Therefore we have that:

(8)

\begin{align} \quad p'(\lambda) = k(\lambda - \lambda)^{k-1}q(\lambda) + (\lambda - \lambda)^k q'(\lambda) \\ \quad p'(\lambda) = 0 \end{align}

Therefore $\lambda$ is a root of $p'$ and so $p$ and $p'$ have a room $\lambda$ in common.