# Sums and Products of Roots

> Find the sum and product of the roots straight from the coefficients, for quadratics, cubics and quartics.

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

### Story time: The detective's clue
A detective is told: "Two numbers add to 10 and multiply to 21." She does not need to try every pair. She writes $x^2 - 10x + 21 = 0$, factorises it to $(x - 3)(x - 7)$, and has her answer: 3 and 7.

That clue works backwards too. The coefficients of a polynomial secretly hold the sum and product of its roots. You can find them without ever solving the equation.

**In short:** The roots of a polynomial are linked to its coefficients. For $ax^2 + bx + c = 0$, the roots add to $-\dfrac{b}{a}$ and multiply to $\dfrac{c}{a}$. For a cubic, the sum is $-\dfrac{b}{a}$, the sum of pairs is $\dfrac{c}{a}$ and the product is $-\dfrac{d}{a}$.

## Starting point: what you already know

- Expanding $(x - 3)(x - 7) = x^2 - 10x + 21$.
- Solving a quadratic by factorising or the quadratic formula.
- From the factor theorem: if the roots are $\alpha$ and $\beta$, then $(x - \alpha)$ and $(x - \beta)$ are factors.
- The Greek letters $\alpha$ (alpha), $\beta$ (beta), $\gamma$ (gamma) and $\delta$ (delta) are used as names for roots.

## Quadratics

Suppose $ax^2 + bx + c = 0$ has roots $\alpha$ and $\beta$. Then

$$ax^2 + bx + c = a(x - \alpha)(x - \beta) = a\big(x^2 - (\alpha + \beta)x + \alpha\beta\big)$$

Matching the coefficients:

### Key idea: Quadratic
$$\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}$$

Check with the detective: $x^2 - 10x + 21$ has $a = 1$, $b = -10$, $c = 21$. Sum $= 10$, product $= 21$.

## Cubics

For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha$, $\beta$, $\gamma$, expanding $a(x - \alpha)(x - \beta)(x - \gamma)$ gives:

### Key idea: Cubic
$$\alpha + \beta + \gamma = -\frac{b}{a}$$
$$\alpha\beta + \beta\gamma + \alpha\gamma = \frac{c}{a}$$
$$\alpha\beta\gamma = -\frac{d}{a}$$

The middle one is the sum of the roots taken **two at a time**.

## Quartics

For $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$:

$$\sum \alpha = -\frac{b}{a}, \quad \sum \alpha\beta = \frac{c}{a}, \quad \sum \alpha\beta\gamma = -\frac{d}{a}, \quad \alpha\beta\gamma\delta = \frac{e}{a}$$

### Exam tip: The sign pattern
Go along the coefficients after the first: minus, plus, minus, plus. Each one is divided by $a$.

### Worked example: A cubic
$2x^3 - 4x^2 + 6x - 1 = 0$ has roots $\alpha$, $\beta$, $\gamma$. Here $a = 2$, $b = -4$, $c = 6$, $d = -1$.

- $\alpha + \beta + \gamma = -\dfrac{-4}{2} = 2$
- $\alpha\beta + \beta\gamma + \alpha\gamma = \dfrac{6}{2} = 3$
- $\alpha\beta\gamma = -\dfrac{-1}{2} = \dfrac{1}{2}$

## Building other expressions

Many expressions can be rewritten using sums and products.

- $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
- $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}$
- For a cubic: $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \alpha\gamma)$

### Worked example: Using the cubic above
$\alpha^2 + \beta^2 + \gamma^2 = 2^2 - 2(3) = -2$.

A sum of squares is negative, so the roots cannot all be real. This is a quick way to show some roots are not real numbers.

**Mind map**

```text
Sums and products of roots
  Quadratic
    Sum = -b/a
    Product = c/a
  Cubic
    Sum = -b/a
    Pairs = c/a
    Product = -d/a
  Quartic
    Signs go minus, plus, minus, plus
  Uses
    α² + β² = (α + β)² - 2αβ
    1/α + 1/β = (α + β)/αβ
    Find unknown coefficients
```

### Year 11: Using extra information about the roots
$x^3 - 6x^2 + kx - 6 = 0$ has roots in the ratio $1 : 2 : 3$. Write them as $m$, $2m$, $3m$.

Sum: $6m = 6$, so $m = 1$. The roots are 1, 2, 3.

Then $k = \alpha\beta + \beta\gamma + \alpha\gamma = 2 + 6 + 3 = 11$. Check the product: $1 \times 2 \times 3 = 6 = -\dfrac{-6}{1}$. Good.

**Two roots are opposites.** Write the roots as $\alpha$, $-\alpha$ and $\beta$. The sum is just $\beta$, so the sum formula gives one root at once. For $x^3 - 3x^2 + kx + 12 = 0$: $\beta = 3$, and putting $x = 3$ into the equation gives $27 - 27 + 3k + 12 = 0$, so $k = -4$.

**Using the polynomial itself.** Each root satisfies the equation, so for $x^3 - 2x^2 + x + 5 = 0$, $\alpha^3 = 2\alpha^2 - \alpha - 5$. Adding the three versions gives $\sum\alpha^3 = 2\sum\alpha^2 - \sum\alpha - 15$.

**Products like $(\alpha + 1)(\beta + 1)(\gamma + 1)$.** A monic cubic is $P(x) = (x - \alpha)(x - \beta)(x - \gamma)$, so $P(-1) = -(\alpha + 1)(\beta + 1)(\gamma + 1)$. Work out $P(-1)$ and change the sign.

**Quartic sums of squares.** $\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$ works for any degree.

### Year 11: Making a new equation
The roots of $x^2 - 3x + 1 = 0$ are $\alpha$ and $\beta$. Find a quadratic with roots $2\alpha$ and $2\beta$.

New sum $= 2(\alpha + \beta) = 6$. New product $= 4\alpha\beta = 4$. So the equation is $x^2 - 6x + 4 = 0$.

**Substitution is faster for cubics.** If $\alpha, \beta, \gamma$ are the roots of $P(x) = 0$, then:
- roots $2\alpha, 2\beta, 2\gamma$: solve $P\left(\frac{x}{2}\right) = 0$, because $x = 2\alpha$ means $\frac{x}{2} = \alpha$;
- roots $\alpha + 1, \beta + 1, \gamma + 1$: solve $P(x - 1) = 0$.

For $x^3 - 2x + 3 = 0$, the equation with roots $2\alpha, 2\beta, 2\gamma$ is $\frac{x^3}{8} - x + 3 = 0$, that is $x^3 - 8x + 24 = 0$.

### Year 12: Roots that form a pattern
Some exam questions give roots in an arithmetic sequence, such as $a - d$, $a$, $a + d$. Then the sum is simply $3a$, which finds the middle root at once. This links to the sequences you meet in Mathematics Advanced in Year 12.

**A double root** $\alpha$ is a root of both $P$ and $P'$. Solve $P'(x) = 0$ and test which answer also makes $P(x) = 0$.

## Quick summary

- Quadratic: $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$.
- Cubic: sum $-\dfrac{b}{a}$, pairs $\dfrac{c}{a}$, product $-\dfrac{d}{a}$.
- Quartic: the signs go minus, plus, minus, plus.
- Rewrite other expressions, such as $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

## Quiz

Try these questions from the chapter quiz.

1. Two numbers add to 9 and multiply to 20. They are the roots of:
   - A. $x^2 + 9x + 20 = 0$
   - B. $x^2 - 9x + 20 = 0$
   - C. $x^2 - 20x + 9 = 0$
   - D. $x^2 + 9x - 20 = 0$
   - Answer: B. $x^2 - 9x + 20 = 0$. $x^2 - (\text{sum})x + (\text{product}) = 0$.

2. For $x^2 - 5x + 6 = 0$, find $\alpha + \beta$.
   - A. $-5$
   - B. 5
   - C. 6
   - D. $-6$
   - Answer: B. 5. $-\dfrac{b}{a} = 5$.

3. For $3x^2 + 6x - 2 = 0$, find $\alpha\beta$.
   - A. 2
   - B. $-2$
   - C. $-\dfrac{2}{3}$
   - D. $\dfrac{2}{3}$
   - Answer: C. $-\dfrac{2}{3}$. $\dfrac{c}{a} = -\dfrac{2}{3}$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/sums-and-products-of-roots/#quiz