# Polynomials and the Factor Theorem

> Polynomial language, expanding, long division, the remainder and factor theorems, finding zeros and sketching cubics.

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

### Story time: The box from a sheet of card
Ruby has a 20 cm by 20 cm sheet of card. She cuts a square of side $x$ from each corner and folds up the sides to make an open box. The volume is

$$V = x(20 - 2x)^2 = 4x^3 - 80x^2 + 400x$$

That is a **polynomial**: a sum of powers of $x$. To find which cut gives a box of exactly 500 cm³, she has to solve a cubic equation. The factor theorem is the tool that cracks equations like this open.

**In short:** A polynomial is a sum of terms like $ax^n$ with whole number powers, and its degree is the highest power. By the factor theorem, if $P(a) = 0$ then $(x - a)$ is a factor, and by the remainder theorem dividing $P(x)$ by $(x - a)$ leaves $P(a)$.

## Starting point: Years 9 and 10

### Polynomial words

A **polynomial** is a sum of terms $ax^n$ where each power $n$ is a whole number $0, 1, 2, \ldots$

For $P(x) = 2x^3 - 5x^2 + x - 7$:
- The **degree** is the highest power: 3. So it is a **cubic**.
- The **leading term** is $2x^3$ and the **leading coefficient** is 2.
- The **constant term** is $-7$.

Degree 1 is linear, 2 is quadratic, 3 is cubic, 4 is quartic. $\dfrac{1}{x}$ and $\sqrt{x}$ are not polynomials, because their powers are not whole numbers.

### Evaluating and expanding

$P(2)$ means substitute $x = 2$: $P(2) = 16 - 20 + 2 - 7 = -9$.

To expand, multiply every term by every term and collect like terms:

$$(x + 2)(x^2 - 3x + 1) = x^3 - 3x^2 + x + 2x^2 - 6x + 2 = x^3 - x^2 - 5x + 2$$

### Dividing polynomials

Long division works like number long division: divide the leading terms, multiply back, subtract, bring down, repeat.

### Worked example: Divide
Divide $x^3 + 2x^2 - 5x - 6$ by $x - 2$.

1. $x^3 \div x = x^2$. Multiply: $x^3 - 2x^2$. Subtract: $4x^2 - 5x$.
2. $4x^2 \div x = 4x$. Multiply: $4x^2 - 8x$. Subtract: $3x - 6$.
3. $3x \div x = 3$. Multiply: $3x - 6$. Subtract: $0$.

So $x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3)$, with remainder 0.

### The remainder theorem

When $P(x)$ is divided by $x - a$, the remainder is $P(a)$.

Check: $P(2) = 8 + 8 - 10 - 6 = 0$, matching the remainder above.

### Calculator tip: Evaluate P(a) quickly
To find a remainder, you only need $P(a)$. Type $P(x)$ once with the letter `X`, for example `2X³ - 3X² + X - 5`, then use `CALC` (if your calculator has it) or store $a$ in `X` with `STO`. Testing several values this way is a fast search for a factor: a result of 0 means $(x - a)$ is a factor.

### The factor theorem

### Key idea: Factor theorem
If $P(a) = 0$, then $x - a$ is a factor of $P(x)$. And if $x - a$ is a factor, then $P(a) = 0$.

### Finding all the zeros

1. Try small factors of the constant term: $\pm 1, \pm 2, \pm 3, \ldots$ until $P(a) = 0$.
2. Divide by $x - a$ to get a quadratic.
3. Factorise the quadratic.

From the example, $x^2 + 4x + 3 = (x + 1)(x + 3)$, so

$$x^3 + 2x^2 - 5x - 6 = (x - 2)(x + 1)(x + 3)$$

The zeros are $x = 2$, $-1$ and $-3$.

### Watch out: The sign of the factor
$P(-3) = 0$ means $x + 3$ is a factor, not $x - 3$.

### Cubic graphs

$y = a(x - h)^3 + k$ is $y = x^3$ moved right $h$ and up $k$. Its flat **point of inflection** is at $(h, k)$. A negative $a$ flips it.

A factorised cubic like $y = (x - 2)(x + 1)(x + 3)$ crosses the x-axis at each zero. With a positive leading coefficient it starts low on the left and ends high on the right.

**Mind map**

```text
Polynomials
  Language
    Degree, leading term
    Constant term
  Operations
    Expand
    Long division
  Theorems
    Remainder: P(a)
    Factor: P(a) = 0
  Zeros
    Try factors of constant
    Divide, then factorise
  Graphs
    y = a(x - h)³ + k
    Cross at single zeros
    Touch at double zeros
```

### Year 11: Sketching polynomials
For $y = P(x)$ in factorised form:
- A **single** zero: the graph cuts straight through.
- A **double** zero, from $(x - a)^2$: the graph touches the axis and turns back.
- A **triple** zero, from $(x - a)^3$: it flattens and passes through.
- The y-intercept is $P(0)$.
- The ends follow the leading term. For an even degree with $a > 0$, both ends go up. For an odd degree with $a > 0$, it goes from bottom left to top right.

**Example.** $y = -x(x - 3)^2$: zeros at 0 (cut) and 3 (touch), leading term $-x^3$, so it comes down from top left and finishes bottom right.

**Equations.** A polynomial equation of degree $n$ has at most $n$ real solutions.

**Degree of a product.** Multiplying polynomials adds their degrees: a cubic times a quadratic has degree 5.

**Equating coefficients.** If two polynomials are equal **for all $x$**, the coefficients of each power match. For $3x^2 + 5x + 1 = a(x + 1)^2 + b(x + 1) + c$: expand the right side to $ax^2 + (2a + b)x + (a + b + c)$, so $a = 3$, $2a + b = 5$ gives $b = -1$, and $a + b + c = 1$ gives $c = -1$. (Substituting handy values of $x$, like $x = -1$, also works.)

**Building from the roots.** A monic cubic with roots $-2$, $0$ and $2$ is $y = x(x + 2)(x - 2) = x^3 - 4x$.

**Two unknowns, two factors.** If $x - 1$ and $x + 3$ are both factors of $P(x) = x^3 + ax^2 + bx - 3$, then $P(1) = 0$ and $P(-3) = 0$ give two equations: $a + b = 2$ and $9a - 3b = 30$. Solving gives $a = 3$, $b = -1$.

**Odd and even polynomials.** A polynomial with only odd powers is odd, and with only even powers (and a constant) is even. If an odd polynomial has a root at $x = 2$, it also has one at $x = -2$.

**Watch the leading term.** In $y = x^2(x - 2)(1 - x)$, the factor $(1 - x)$ contributes $-x$, so the leading term is $-x^4$ and both ends go down.

### Year 12: Polynomials in calculus
Polynomials are the easiest functions to differentiate and integrate, so they appear in many calculus questions.

Ruby's box: $V = 4x^3 - 80x^2 + 400x$ for $0 < x < 10$. Then

$$\frac{dV}{dx} = 12x^2 - 160x + 400 = 4(3x - 10)(x - 10)$$

which is zero at $x = \dfrac{10}{3}$ (inside the domain). So the largest box has a cut of about 3.33 cm and volume about 593 cm³. The factorising skills of this chapter find stationary points quickly.

## Quick summary

- A polynomial is a sum of terms $ax^n$ with whole number powers. The degree is the highest power.
- Long division: divide leading terms, multiply, subtract, repeat.
- Remainder theorem: dividing by $x - a$ leaves $P(a)$.
- Factor theorem: $P(a) = 0$ means $x - a$ is a factor.
- Find zeros by testing factors of the constant term, dividing, then factorising.

## Quiz

Try these questions from the chapter quiz.

1. What is the degree of $5x^4 - 2x^2 + 7$?
   - A. 2
   - B. 4
   - C. 5
   - D. 7
   - Answer: B. 4. The highest power is 4.

2. Which is NOT a polynomial?
   - A. $x^3 - 1$
   - B. $7$
   - C. $x^2 + \dfrac{1}{x}$
   - D. $3x - 2$
   - Answer: C. $x^2 + \dfrac{1}{x}$. $\dfrac{1}{x} = x^{-1}$ has a negative power.

3. What is the leading coefficient of $-3x^3 + x - 4$?
   - A. $-3$
   - B. 3
   - C. $-4$
   - D. 1
   - Answer: A. $-3$. The number in front of the highest power.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/polynomials/#quiz