Multiple zeroes and the derivative

A zero of multiplicity m of a polynomial is a zero of multiplicity m − 1 of its derivative, testing for double zeroes with P(α) = P'(α) = 0, and finding unknown coefficients from a multiple zero.

The theorem

Example 1: proof for a double zero●●●

Suppose $P(x) = (x - \alpha)^2 Q(x)$. Show that $P'(\alpha) = 0$.

By the product rule, $P'(x) = 2(x - \alpha)Q(x) + (x - \alpha)^2 Q'(x) = (x - \alpha)\big(2Q(x) + (x - \alpha)Q'(x)\big)$.

So $(x - \alpha)$ is a factor of $P'(x)$, and $P'(\alpha) = 0$.

Finding unknown coefficients

Example 2●●

$P(x) = x^3 - 3x^2 + k$ has a double zero, with $k \ne 0$. Find $k$ and factor $P(x)$.

$P'(x) = 3x^2 - 6x = 3x(x - 2)$, so a double zero is at $x = 0$ or $x = 2$.

$x = 0$ would need $k = 0$. So the double zero is at $x = 2$: $P(2) = 8 - 12 + k = 0$, giving $k = 4$.

Then $P(x) = x^3 - 3x^2 + 4 = (x - 2)^2(x + 1)$.

Example 3●●●

Find $a$ and $b$ if $x = 1$ is a double zero of $P(x) = x^3 + ax + b$.

$P(1) = 1 + a + b = 0$ and $P'(1) = 3 + a = 0$.

So $a = -3$ and $b = 2$, giving $P(x) = x^3 - 3x + 2 = (x - 1)^2(x + 2)$.

Flashcards

Test for a double zero at $lpha$

$P(\alpha) = 0$ and $P'(\alpha) = 0$.

Zero of multiplicity $m$ of $P$

A zero of multiplicity $m - 1$ of $P'$.

Triple zero test

$P(\alpha) = P'(\alpha) = P''(\alpha) = 0$

Practice

$P(x) = x^3 - 3x^2 + k$ has a double zero and $k \ne 0$. Find $k$.
$x = 1$ is a double zero of $x^3 + ax + b$. Find $a$.
For the same polynomial, find $b$.
$x = 5$ is a zero of multiplicity $3$ of $P(x)$. What is its multiplicity as a zero of $P'(x)$?
$P(x) = x^3 - 3x + 2 = (x - 1)^2(x + c)$. Find the single zero of $P(x)$.
$x = 2$ is a double zero of $P(x) = x^3 + px + 16$. Find $p$.
$P(\alpha) = 0$ and $P'(\alpha) = 0$. Then $\alpha$ is a zero of $P$ of multiplicity:
$P(x) = (x - 3)^4 Q(x)$, where $Q(3) \ne 0$. Then $P''(3)$ is: