Multiplicity of zeroes and the derivative: a zero of multiplicity m is a zero of P'(x) of multiplicity m - 1, finding repeated roots with P and P', proving the result, and linking multiplicity to the shape of the graph (Moved to Year 12 in the 2024 Extension 1 syllabus)

Product rule. With $u = (x - 2)^3$ and $v = x + 1$,

The factor $(x - 2)^2$ shows that $x = 2$ is a zero of $P'(x)$ of multiplicity $2$, one less than its multiplicity $3$ in $P(x)$.

Marker's note: one mark for the derivative, one for the factorisation and the multiplicity statement.

A double zero at $x = 2$ means $P(2) = 0$ and $P'(2) = 0$.

$P(2) = 8 + 4a + 2b + 4 = 0$, so $4a + 2b = -12$.

$P'(x) = 3x^2 + 2ax + b$, so $P'(2) = 12 + 4a + b = 0$, giving $4a + b = -12$.

Solve. Subtracting the second equation from the first gives $b = 0$, and then $a = -3$.

Factorise. $P(x) = x^3 - 3x^2 + 4$. Since $(x - 2)^2$ is a factor, divide to find the other factor: $P(x) = (x - 2)^2 (x + 1)$.

Marker's note: one mark for each of the two conditions, one for the values and the factorisation. Check by expanding $(x - 2)^2(x + 1) = (x^2 - 4x + 4)(x + 1) = x^3 - 3x^2 + 4$.

Test the polynomial and its derivatives at $x = -1$.

$P(-1) = 1 - 2 + 2 - 1 = 0$.

$P'(x) = 4x^3 + 6x^2 - 2$, so $P'(-1) = -4 + 6 - 2 = 0$.

$P''(x) = 12x^2 + 12x$, so $P''(-1) = 12 - 12 = 0$.

$P'''(x) = 24x + 12$, so $P'''(-1) = -12 \neq 0$.

The first nonzero derivative at $x = -1$ is the third, so the multiplicity is $3$.

Factorise. $P(x) = (x + 1)^3 (x + c)$ and comparing the constant terms gives $c = -1$:

Marker's note: one mark for $P$ and $P'$, one for $P''$ and the nonzero $P'''$, one for the factorisation. Expanding $(x + 1)^3(x - 1)$ confirms $x^4 + 2x^3 - 2x - 1$.

A repeated zero must also be a zero of $P'(x)$. $P'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$, so the only candidates are $x = 1$ and $x = -1$.

Substitute each candidate into $P$.

$P(1) = 1 - 3 + k = k - 2 = 0$ gives $k = 2$.

$P(-1) = -1 + 3 + k = k + 2 = 0$ gives $k = -2$.

Therefore $k = 2$ or $k = -2$. Check: $x^3 - 3x + 2 = (x - 1)^2(x + 2)$ and $x^3 - 3x - 2 = (x + 1)^2(x - 2)$.

Marker's note: one mark for the derivative and its zeroes, one for each value of $k$.

Differentiate with the product rule.

Factor out the common power.

Show the remaining factor is nonzero at $\alpha$. Let $R(x) = mQ(x) + (x - \alpha)Q'(x)$. Then $R(\alpha) = mQ(\alpha) + 0 = mQ(\alpha) \neq 0$, because $m \neq 0$ and $Q(\alpha) \neq 0$.

Hence $(x - \alpha)^{m-1}$ divides $P'(x)$ but $(x - \alpha)^m$ does not (the cofactor $R$ does not vanish at $\alpha$), so $\alpha$ is a zero of $P'(x)$ of multiplicity exactly $m - 1$.

Marker's note: one mark for the product rule, one for the factorisation, one for $R(\alpha) = mQ(\alpha)$, one for the conclusion that mentions why $R(\alpha) \neq 0$. Omitting that justification loses the last mark.

(a) Differentiate. $P'(x) = 4x^3 - 6x^2 - 6x + 4 = 2\left( 2x^3 - 3x^2 - 3x + 2 \right)$. Test the three claimed zeroes: at $x = 2$, $16 - 12 - 6 + 2 = 0$; at $x = -1$, $-2 - 3 + 3 + 2 = 0$; at $x = \frac{1}{2}$, $\frac{1}{4} - \frac{3}{4} - \frac{3}{2} + 2 = 0$. A cubic has at most three zeroes, and the leading coefficient $2$ gives $P'(x) = 2(x - 2)(x + 1)(2x - 1)$.

(b) Which zeroes of $P'$ are also zeroes of $P$? A repeated zero of $P$ must be a zero of $P'$. Evaluating,

$P(2) = 16 - 16 - 12 + 8 + 4 = 0$ and $P(-1) = 1 + 2 - 3 - 4 + 4 = 0$, while $P\left( \tfrac{1}{2} \right) = \frac{81}{16} \neq 0$.

So $x = 2$ and $x = -1$ are repeated zeroes. Since $P$ has degree $4$, each is a double zero and

(c) Graph. The graph touches the $x$-axis (without crossing) at $x = -1$ and $x = 2$, where it has local minimum turning points $(-1, 0)$ and $(2, 0)$. Between them there is a local maximum at $x = \frac{1}{2}$ with $y = \frac{81}{16}$. The $y$-intercept is $P(0) = 4$. As $x \to \pm\infty$, $y \to +\infty$.

Marker's note: one mark for $P'$, one for the factorisation, one for testing each zero of $P'$ in $P$, one for the factored form, one for the graph features.

Let $P(x) = x^3 - 3x^2 + m$

A repeated root must satisfy $P'(x) = 3x^2 - 6x = 3x(x - 2) = 0$, so $x = 0$ or $x = 2$.

Substitute

$P(0) = m = 0$ gives $m = 0$. $P(2) = 8 - 12 + m = m - 4 = 0$ gives $m = 4$.

Factorise

For $m = 0$: $P(x) = x^3 - 3x^2 = x^2(x - 3)$. For $m = 4$: $P(x) = x^3 - 3x^2 + 4 = (x - 2)^2(x + 1)$.

Marker's note: one mark for the derivative and its zeroes, one for each value of $m$, one for the factorisations (the second repeated root is double, not triple, because $P''$ is nonzero there).