# Polynomials, Division and the Factor Theorem

> The language of polynomials, long division, and the remainder and factor theorems for finding roots fast.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/polynomial-division/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The engineer's shortcut
Priya designs roller coaster tracks. Each section of track follows a polynomial curve, such as $P(x) = x^3 - 6x^2 + 11x - 6$. She needs to know where the track meets ground level, which means solving $P(x) = 0$.

There is no simple formula like the quadratic formula for most cubics. But Priya knows a shortcut: test easy numbers like 1, 2 and 3. If $P(1) = 0$, then $(x - 1)$ is a factor, and the rest is just division. That shortcut is the **factor theorem**.

**In short:** Dividing a polynomial gives $P(x) = A(x)Q(x) + R(x)$, where the remainder has a lower degree than the divisor. The remainder theorem says dividing by $(x - a)$ leaves $P(a)$, and the factor theorem says $(x - a)$ is a factor exactly when $P(a) = 0$.

## Starting point: what you already know

- A **quadratic** $ax^2 + bx + c$ has a highest power of 2.
- **Expanding:** $(x - 1)(x + 3) = x^2 + 2x - 3$.
- **Factorising** a quadratic and solving $ax^2 + bx + c = 0$.
- **Substitution:** for $P(x) = x^2 + 1$, $P(2) = 5$.
- Long division of numbers: $47 \div 5 = 9$ remainder 2, so $47 = 5 \times 9 + 2$.

## The language of polynomials

A **polynomial** is a sum of terms with whole number powers of $x$:

$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$

| Word | Meaning | In $P(x) = 2x^3 - 5x + 7$ |
|---|---|---|
| degree | highest power | 3 |
| leading term | term with the highest power | $2x^3$ |
| leading coefficient | number in front of it | 2 |
| constant term | term with no $x$ | 7 |
| monic | leading coefficient is 1 | not monic |

A **root** or **zero** of $P$ is a number $\alpha$ with $P(\alpha) = 0$. On the graph, it is where the curve meets the x-axis.

Expressions like $\sqrt{x}$ or $\dfrac{1}{x}$ are **not** polynomials, because the powers are not whole numbers.

## Dividing polynomials

Just like $47 = 5 \times 9 + 2$, any polynomial can be divided by another:

$$P(x) = A(x) \times Q(x) + R(x)$$

where $Q(x)$ is the **quotient** and $R(x)$ is the **remainder**. The remainder has a lower degree than the divisor $A(x)$.

### Worked example: Divide x³ − 2x² + 4x − 5 by x − 1
Work like number long division, one leading term at a time.

1. $x^3 \div x = x^2$. Multiply: $x^2(x - 1) = x^3 - x^2$. Subtract: $-x^2 + 4x$.
2. $-x^2 \div x = -x$. Multiply: $-x(x - 1) = -x^2 + x$. Subtract: $3x - 5$.
3. $3x \div x = 3$. Multiply: $3(x - 1) = 3x - 3$. Subtract: $-2$.

So $x^3 - 2x^2 + 4x - 5 = (x - 1)(x^2 - x + 3) - 2$. Quotient $x^2 - x + 3$, remainder $-2$.

## The remainder theorem

Put $x = a$ into $P(x) = (x - a)Q(x) + R$. The first part becomes zero, so:

### Key idea: Remainder theorem
When $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.

Check the example: $P(1) = 1 - 2 + 4 - 5 = -2$. Same remainder, no division needed.

### Calculator tip: Evaluate P(a) quickly
Type $P(x)$ once with the letter `X`, then use `CALC` (if your calculator has it) or store each value in `X`. A result of 0 means $(x - a)$ is a factor, so trying $\pm 1$, $\pm 2$ and so on is fast.

## The factor theorem

If the remainder is zero, $(x - a)$ divides exactly.

### Key idea: Factor theorem
$(x - a)$ is a factor of $P(x)$ exactly when $P(a) = 0$.

### Worked example: Factorise P(x) = x³ − 6x² + 11x − 6
Try factors of the constant term 6: $\pm 1, \pm 2, \pm 3, \pm 6$.

$P(1) = 1 - 6 + 11 - 6 = 0$, so $(x - 1)$ is a factor. Divide to get $x^2 - 5x + 6 = (x - 2)(x - 3)$.

So $P(x) = (x - 1)(x - 2)(x - 3)$ and the roots are 1, 2 and 3. Priya's track touches the ground three times.

### Exam tip: Where to look for roots
For a polynomial with whole number coefficients, any whole number root must divide the constant term. Test those first.

**Mind map**

```text
Polynomials
  Language
    Degree and leading term
    Monic, constant term
    Roots are zeros
  Division
    P = A × Q + R
    Remainder has lower degree
  Remainder theorem
    Remainder is P(a)
  Factor theorem
    P(a) = 0 means x - a is a factor
    Test factors of the constant
```

### Year 11: Finding unknown coefficients
$P(x) = x^3 + ax^2 - 4x + b$ has a factor $(x - 2)$ and leaves remainder 6 when divided by $(x + 1)$. Find $a$ and $b$.

- $P(2) = 0$: $8 + 4a - 8 + b = 0$, so $4a + b = 0$.
- $P(-1) = 6$: $-1 + a + 4 + b = 6$, so $a + b = 3$.

Subtract: $3a = -3$, so $a = -1$ and $b = 4$.

### Year 11: Dividing by a quadratic
When dividing by a quadratic, the remainder can be linear, $R(x) = px + q$.

**Example.** Find the remainder when $P(x)$ is divided by $(x - 1)(x - 2)$, given $P(1) = 3$ and $P(2) = 5$. Write $P(x) = (x - 1)(x - 2)Q(x) + px + q$. Then $p + q = 3$ and $2p + q = 5$, so $p = 2$, $q = 1$. The remainder is $2x + 1$.

**Long division by a quadratic** works the same way, one leading term at a time. $x^3 + 3x^2 + 5 \div (x^2 + 2)$: $x^3 \div x^2 = x$, subtract $x^3 + 2x$ to leave $3x^2 - 2x + 5$; then $3x^2 \div x^2 = 3$, subtract $3x^2 + 6$ to leave $-2x - 1$. So the quotient is $x + 3$ and the remainder $-2x - 1$.

**Using values you already know.** The remainder on dividing $x^{8}$ by $x^2 - 1$ is $px + q$ with $1 = p + q$ (at $x = 1$) and $1 = -p + q$ (at $x = -1$), so the remainder is 1. If dividing by $x^2 - 9$ leaves $2x + 5$, then dividing by $x - 3$ leaves $P(3) = 11$, because the quadratic part is zero at $x = 3$.

**Odd and even polynomials.** An odd polynomial has only odd powers ($P(x) = ax^3 + cx$ for a cubic); an even one has only even powers. Use this to cut down the unknowns before using the factor theorem.

**Unknown inside the factor.** If $x + k$ is a factor, then $P(-k) = 0$. Factorise the resulting equation in $k$ and discard any value the question rules out.

### Year 12: Polynomials and calculus
In Year 12 you differentiate polynomials to find turning points and use them to model motion. A double root of $P$ is also a root of $P'$, which links this chapter to calculus.

**Remainder on dividing by $(x - a)^2$.** Write $P(x) = (x - a)^2Q(x) + rx + s$. Then $P(a) = ra + s$, and differentiating gives $P'(a) = r$. For $P(x) = x^3 + 2x$ divided by $(x - 1)^2$: $P(1) = 3$ and $P'(1) = 5$, so $r = 5$, $s = -2$, and the remainder is $5x - 2$.

**Finding a double root.** Solve $P'(x) = 0$, then test those values in $P$. For $x^3 - 3x + 2$: $P'(x) = 3x^2 - 3 = 0$ at $x = \pm 1$, and $P(1) = 0$, so $(x - 1)^2$ is a factor: $x^3 - 3x + 2 = (x - 1)^2(x + 2)$. For an unknown coefficient, solve $P(\alpha) = 0$ and $P'(\alpha) = 0$ together.

**Conditions like "$P(x) - 1$ is divisible by $(x - 1)^2$"** mean $P(1) = 1$ and $P'(1) = 0$.

## Quick summary

- A polynomial has whole number powers. The degree is the highest power.
- $P(x) = A(x)Q(x) + R(x)$, with the remainder of lower degree than $A(x)$.
- Remainder theorem: dividing by $(x - a)$ leaves remainder $P(a)$.
- Factor theorem: $(x - a)$ is a factor exactly when $P(a) = 0$.
- To factorise a cubic, test factors of the constant term, then divide.

## Quiz

Try these questions from the chapter quiz.

1. What is the degree of $P(x) = 4x^5 - x^2 + 9$?
   - A. 4
   - B. 5
   - C. 9
   - D. 2
   - Answer: B. 5. The highest power of $x$ is 5.

2. What is the leading coefficient of $3 - 2x + 7x^3$?
   - A. 3
   - B. $-2$
   - C. 7
   - D. 1
   - Answer: C. 7. The highest power term is $7x^3$.

3. Which is a monic polynomial?
   - A. $2x^2 + 1$
   - B. $x^3 - 4x + 2$
   - C. $5x^4 + x$
   - D. $-x^2 + 3$
   - Answer: B. $x^3 - 4x + 2$. Monic means the leading coefficient is 1.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/polynomial-division/#quiz