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Module 2 · L11 of 15 ~30 min ⚡ +90 XP available

Multiple Roots

When you factorise a polynomial, some factors appear more than once, like $(x-2)^3$ or $(x+1)^2$. This repetition is called multiplicity, and it completely controls whether the graph crosses or merely touches the $x$-axis. Master multiplicity and sketching polynomials becomes a matter of reading the factored form.

Today's hook, The graph of $y = (x-2)^2$ never goes below the $x$-axis, it bounces off at $x=2$. But $y=(x-2)^3$ slices right through the axis at the same point. Same root, completely different graph behaviour. The only difference is whether the exponent is even or odd. By the end of this lesson you'll know exactly why, and be able to read graph shape directly from a factored polynomial.

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Recall, your gut answer first

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What is the difference between a root of $P(x) = 0$ and a factor of $P(x)$? Without looking ahead if $(x - 2)^2$ is a factor of $P(x)$, how many times does $x = 2$ appear as a root, and what do you think this means for the graph?

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Orientation and key terms

The one rule that governs everything

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The entire lesson comes down to one idea: the multiplicity of a root is the exponent of its factor, and whether that multiplicity is even or odd tells you everything about the graph's behaviour at that intercept.

For a root $\alpha$ of $P(x) = a(x-\alpha)^m \cdot Q(x)$ where $Q(\alpha) \neq 0$:

If $m$ is odd: the sign of $P(x)$ changes across $\alpha$, so the graph crosses the $x$-axis.
If $m$ is even: the sign of $P(x)$ does not change across $\alpha$, so the graph touches the $x$-axis and turns back.

y = (x+2)(x−1)² crosses the axis at the odd root x = −2 and touches at the even root x = 1.

$\text{odd } m \Rightarrow \text{cross} \quad \text{even } m \Rightarrow \text{touch}$

Read the exponent

Multiplicity = the exponent of $(x - \alpha)$ in the fully factored form. No factoring needed after that, just check odd vs even.

Degree check

The sum of all multiplicities equals the degree of the polynomial. Use this as a quick sanity-check when factoring.

Why it works

$(x-\alpha)^m$ stays positive if $m$ is even (squared terms can't go negative), so the function can't change sign, it bounces.

What you'll master

Know

Key facts

The definition of multiplicity: the exponent of the corresponding factor.
A root with multiplicity $m$ contributes $m$ to the total degree.
Odd multiplicity roots cross the axis; even multiplicity roots touch it.

Understand

Concepts

Why sign change (or lack of it) at a root depends on the parity of multiplicity.
How multiplicity affects the "flatness" of the graph near the root, higher multiplicity is flatter.
The connection between roots, factors, and the degree of a polynomial.

Can do

Skills

State the multiplicity of each root from a factored polynomial.
Determine whether the graph crosses or touches at each $x$-intercept.
Find all roots and their multiplicities by factoring $P(x)$.

Key terms

MultiplicityThe number of times a root appears; the exponent of its corresponding linear factor in the factored form.

Repeated rootA root with multiplicity greater than 1, e.g. $x = 2$ in $(x-2)^3$.

Simple rootA root with multiplicity exactly 1. The graph crosses the axis sharply, like a straight line, at this point.

Double rootA root with multiplicity 2. The graph touches the axis and turns around, it behaves like a parabola vertex near this root.

Point of inflection rootA root with multiplicity 3. The graph crosses but is flat (zero gradient) at the intercept, called a horizontal point of inflection.

Degree (total)Sum of all multiplicities across all roots equals the degree of the polynomial.

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Multiplicity and graph behaviour

core concept

Given a polynomial with a root $\alpha$ of multiplicity $m$, we can write $P(x) = (x - \alpha)^m \cdot Q(x)$ where $Q(\alpha) \neq 0$. The sign of $(x - \alpha)^m$ near $x = \alpha$ determines graph behaviour:

$m$ odd: $(x - \alpha)^m$ changes sign across $\alpha$ (negative one side, positive the other). The graph crosses the axis. The larger the odd $m$, the flatter the crossing, it looks like $y = x^3$ rather than $y = x$ at that point.
$m$ even: $(x - \alpha)^m \geq 0$ always (a non-negative quantity). So $P(x)$ cannot change sign at $\alpha$. The graph touches the axis and returns to the same side. The curve looks like a parabola ($m = 2$) or flatter ($m = 4$) at that root.

$$P(x) = (x-\alpha)^m \cdot Q(x), \quad Q(\alpha) \neq 0$$ $$m \text{ odd} \Rightarrow \text{crosses at } \alpha \qquad m \text{ even} \Rightarrow \text{touches at } \alpha$$

Examples:

$y = (x - 1)^3$: root at $x = 1$ with multiplicity 3 (odd), crosses, with a horizontal point of inflection.
$y = (x - 2)^2(x + 1)$: double root at $x = 2$ (even), touches; simple root at $x = -1$ (odd), crosses.
$y = x^3(x+2)^2$: triple root at $x = 0$ (odd), crosses flatly; double root at $x = -2$ (even), touches.

Degree check. For $y = (x-2)^2(x+1)$, the total degree is $2 + 1 = 3$. Always verify: sum of multiplicities = degree of polynomial. A degree-4 polynomial could have multiplicities (4), (3,1), (2,2), (2,1,1), (1,1,1,1), all must add to 4.

Multiplicity = exponent of the factor (x - ) in fully factored form; Odd multiplicity graph crosses x-axis at that root

Pause, copy the multiplicity-behaviour rules into your book: odd multiplicity at root $\alpha$ → graph crosses $x$-axis; even multiplicity → graph touches (bounces off) the $x$-axis at $\alpha$.

Quick check: For $P(x) = (x+1)^4(x-3)$, the behaviour at $x = -1$ is:

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · READING MULTIPLICITY

State the multiplicity of each root of $P(x) = (x - 1)^2(x + 3)^3(x - 5)$ and the total degree.

Root $x = 1$: exponent of $(x-1)$ is $2$ → multiplicity $\mathbf{2}$

Read the exponent directly from the factor $(x-1)^2$.

PROBLEM 2 · FINDING ROOTS BY FACTORING

Find all roots of $P(x) = x^4 - 5x^2 + 4$ and state their multiplicities.

Let $u = x^2$: $u^2 - 5u + 4 = (u-1)(u-4) = 0$

Recognise the hidden quadratic pattern: only even powers of $x$, so substitute $u = x^2$.

PROBLEM 3 · REPEATED ROOTS FROM FACTORING

Find all roots of $x^5 - x^3 = 0$ and state their multiplicities.

$x^5 - x^3 = x^3(x^2 - 1) = x^3(x-1)(x+1)$

Factor out the highest power of $x$ first, then apply difference of squares to $x^2 - 1$.

Did you get this? True or false: the graph of $y = (x-2)^2(x+1)$ touches the $x$-axis at $x = 2$ and crosses at $x = -1$.

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Misconceptions to fix

Misconceptions to fix · the traps that cost marks

Trap 01

Confusing "touches" with "does not cross"

Students sometimes write "a double root means the graph doesn't touch the axis." Wrong, it absolutely touches the axis (the graph hits zero at that $x$ value). The point is that it doesn't cross it bounces back. The graph reaches zero but returns to the same side.

Trap 02

Forgetting multiplicity 3 is a crossing

A triple root looks like a turning point near the intercept but is actually a crossing. The graph is flat at the axis (horizontal point of inflection) but does cross from one side to the other. It behaves like $y = x^3$ at that point, not $y = x^2$.

Trap 03

Not identifying the root correctly from the factor

The factor $(x + 3)$ gives root $x = -3$, not $x = +3$. The root is the value that makes the factor zero. Always set the factor equal to zero: $x + 3 = 0 \Rightarrow x = -3$.

Fill the gap: For $P(x) = (x-1)^2(x+3)^3(x-5)$, the sum of all multiplicities equals . The graph touches the axis at $x = $ .

Did you get this? True or false: the graph of $y = (x-3)^3$ crosses the $x$-axis at $x = 3$.

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Activities

Activities · practice with the ideas

State the multiplicity of each root and whether the graph crosses or touches the $x$-axis at each root for $P(x) = (x-2)^2(x+1)(x-4)^3$.

Find all roots and their multiplicities for $P(x) = x^4 - 8x^2 + 16$.

A degree-5 polynomial has roots $x = 1$ (multiplicity 2), $x = -3$ (multiplicity 2), and $x = 4$ (multiplicity 1). Write it in factored form with leading coefficient 1.

For the polynomial $Q(x) = x^6 - x^4$, factor completely and state the multiplicity of each root.

Explain why an even-multiplicity root causes the graph to bounce off the $x$-axis rather than cross through it.

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Revisit your thinking

Earlier you were asked: if $(x-2)^2$ is a factor, how many times does $x = 2$ appear as a root?

The answer is that $x = 2$ appears as a root with multiplicity 2. It contributes $2$ to the total degree, and because $2$ is even, the graph touches the axis at $x = 2$ rather than crossing through it. The factor $(x-2)^2$ is always $\geq 0$, so the polynomial cannot change sign there, the graph bounces back.

Why does even multiplicity cause the "bounce"? Because raising any factor to an even power removes the ability to change sign. $(x-2)^2 = 0$ at $x=2$, but just to either side it is positive, so the function hugs the axis from one side only.

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Odd one out: Which of these roots does not cause the graph to cross the $x$-axis?

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Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

ApplyBand 32 marks

Q1. State the multiplicity of each root of $P(x) = (x - 1)^2(x + 3)^3(x - 5)$. (2 marks)

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ApplyBand 43 marks

Q2. For $y = (x - 2)^2(x + 1)$: (a) identify the $x$-intercepts and the behaviour of the graph at each; (b) find the $y$-intercept. (3 marks)

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AnalyseBand 42 marks

Q3. Find all roots of $x^5 - x^3 = 0$ and state their multiplicities. (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $x=2$: mult 2, touches; $x=-1$: mult 1, crosses; $x=4$: mult 3, crosses (flat) · 2. $u=x^2$: $u^2-8u+16=(u-4)^2$, so $x^2=4$, $x=\pm2$, both with multiplicity 2 · 3. $P(x)=(x-1)^2(x+3)^2(x-4)$ · 4. $Q(x)=x^4(x^2-1)=x^4(x-1)(x+1)$; $x=0$ mult 4, $x=1$ mult 1, $x=-1$ mult 1 · 5. $(x-\alpha)^m$ with $m$ even is always $\geq 0$, so $P(x)$ keeps the same sign on both sides of $\alpha$, it cannot cross the axis.

Q1 (2 marks): $x=1$ mult 2 [0.5]; $x=-3$ mult 3 [0.5]; $x=5$ mult 1 [0.5]; total degree $=6$ [0.5].

Q2 (3 marks): (a) $x$-intercepts: $x=2$ (mult 2, even, touches); $x=-1$ (mult 1, odd, crosses) [2]. (b) $y = (0-2)^2(0+1) = 4$ [1].

Q3 (2 marks): $x^5-x^3 = x^3(x-1)(x+1)$ [1]. Roots: $x=0$ mult 3, $x=1$ mult 1, $x=-1$ mult 1 [1].

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Review and finish

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Key takeaways

Multiplicity = exponent of the corresponding factor in fully factored form.
Odd multiplicity → graph crosses the $x$-axis at that root.
Even multiplicity → graph touches the $x$-axis and turns around.
Multiplicity 3 (odd): crosses with a flat horizontal point of inflection.
Sum of all multiplicities = degree of polynomial.

Next lesson: Graphing Polynomials combining multiplicity, end behaviour, and intercepts to sketch polynomial graphs.

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