# Roots of polynomial equations

> CIE A-Level Further Mathematics · Further Pure 1
> Source: https://www.owlsprep.com/study/cie-9231-u1-roots-of-polynomial-equations/

This sub-topic covers Vieta's formulae for polynomials up to degree 4, how to calculate symmetric functions of roots, and how to construct new polynomials with transformed roots, a core skill for further pure problems.

**Prerequisites:** [Polynomial basics and algebraic manipulation](https://www.owlsprep.com/study/cie-9231-u0-polynomial-basics/); [Quadratic equations and roots](https://www.owlsprep.com/study/alevel-maths-quadratic-equations/)

## Learning objectives

- Recall and apply Vieta's formulae for polynomials of degree 2, 3 and 4
- Express symmetric functions of roots in terms of fundamental symmetric sums
- Construct new polynomials with roots that are transformations of original roots
- Solve common exam problems involving roots of polynomials without factoring

## Vieta's Formulae for Polynomials

For any polynomial of degree $n$, we can relate the polynomial's coefficients to sums and products of its roots without calculating the roots explicitly. This set of relations is known as **Vieta's formulae**.

**Vieta's Formulae (Monic Polynomials)** — For a monic polynomial with roots $\alpha_1, \alpha_2, ..., \alpha_n$, Vieta's formulae state that the elementary symmetric sums of roots are related to coefficients with alternating signs.

*Notation:* P(x) = x^n + a_{n-1}x^{n-1} + ... + a_0

$$\begin{aligned} \text{Degree 2: } &x^2 + ax + b = 0, \quad \alpha+\beta = -a, \quad \alpha\beta = b \\ \text{Degree 3: } &x^3 + ax^2 + bx + c = 0, \quad \sum\alpha = -a, \quad \sum\alpha\beta = b, \quad \alpha\beta\gamma = -c \\ \text{Degree 4: } &x^4 + ax^3 + bx^2 + cx + d = 0, \quad \sum\alpha = -a, \quad \sum\alpha\beta = b, \quad \sum\alpha\beta\gamma = -c, \quad \alpha\beta\gamma\delta = d \end{aligned}$$

**Worked example:** The cubic $2x^3 - 6x^2 + 4x - 1 = 0$ has roots $\alpha, \beta, \gamma$. Find $\alpha+\beta+\gamma$, $\alpha\beta + \alpha\gamma + \beta\gamma$ and $\alpha\beta\gamma$.

1. First, divide the entire polynomial by the leading coefficient 2 to make it monic:
2. $$x^3 - 3x^2 + 2x - \frac{1}{2} = 0$$
3. Match to the general monic form $x^3 + ax^2 + bx + c = 0$, so $a=-3$, $b=2$, $c=-\frac{1}{2}$
4. Apply Vieta's formula:
5. $$\alpha+\beta+\gamma = -a = 3, \quad \alpha\beta + \alpha\gamma + \beta\gamma = b = 2, \quad \alpha\beta\gamma = -c = \frac{1}{2}$$

> **tip**
>
> Always check if your polynomial is monic before applying the standard formula. Failure to adjust for a leading coefficient not equal to 1 is the most common mistake in this topic.

> **Exam tip:** Always state you are using Vieta's formula in your working to earn method marks.

## Calculating Symmetric Functions of Roots

Any symmetric function of the roots of a polynomial can be expressed in terms of the elementary symmetric sums we get from Vieta's formula. This lets us calculate values of these functions without finding the roots themselves.

**Symmetric Function** — A function of multiple roots that remains unchanged when any two roots are swapped. All symmetric functions can be written using the elementary symmetric sums from Vieta.

**Worked example:** For the cubic $2x^3 - 6x^2 + 4x - 1 = 0$ from the previous example, find $\alpha^2 + \beta^2 + \gamma^2$ and $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$. We already know $\alpha+\beta+\gamma=3$, $\sum \alpha\beta=2$, $\alpha\beta\gamma=\frac{1}{2}$.

1. Use the standard identity for sum of squares:
2. $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$$
3. Substitute the known values:
4. $$= 3^2 - 2(2) = 9 - 4 = 5$$
5. For the sum of reciprocals, combine the fractions:
6. $$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma} = \frac{2}{1/2} = 4$$

## Forming Polynomials with Transformed Roots

The most common exam question in this topic asks you to find a new polynomial whose roots are a function of the roots of an original polynomial. The substitution method is the fastest and most reliable approach for this.

**Worked example:** The cubic equation $x^3 - 4x^2 + 3x - 2 = 0$ has roots $\alpha, \beta, \gamma$. Find a cubic equation with integer coefficients whose roots are $2\alpha+1, 2\beta+1, 2\gamma+1$.

1. Let $y$ be the new root, so $y = 2x + 1$ where $x$ is an original root. Rearrange to get $x$ in terms of $y$:
2. $$x = \frac{y - 1}{2}$$
3. Substitute into the original equation, since $x$ satisfies the original polynomial:
4. $$\left(\frac{y - 1}{2}\right)^3 - 4\left(\frac{y - 1}{2}\right)^2 + 3\left(\frac{y - 1}{2}\right) - 2 = 0$$
5. Multiply through by 8 to eliminate all denominators:
6. $$(y - 1)^3 - 16(y - 1)^2 + 12(y - 1) - 16 = 0$$
7. Expand and collect like terms:
8. $$y^3 - 3y^2 + 3y - 1 - 16y^2 + 32y - 16 + 12y - 12 - 16 = 0$$
9. $$y^3 - 19y^2 + 47y - 45 = 0$$

> **tip**
>
> For linear transformations of roots, the substitution method is faster than calculating each symmetric sum separately for the new polynomial.

> **Exam tip:** Always check your final polynomial has integer coefficients if the question asks for this.

## Sum of Higher Powers of Roots

You may be asked to find the sum of higher powers of roots, which can be done using standard algebraic identities for lower powers, building up to the required power.

**Worked example:** The quadratic equation $x^2 - 5x + 6 = 0$ has roots $\alpha$ and $\beta$. Find $\alpha^4 + \beta^4$.

1. First get the basic symmetric sums from Vieta:
2. $$\alpha + \beta = 5, \quad \alpha\beta = 6$$
3. Calculate the sum of squares first:
4. $$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25 - 12 = 13$$
5. Square this result to get the sum of fourth powers:
6. $$(\alpha^2 + \beta^2)^2 = \alpha^4 + 2\alpha^2\beta^2 \implies \alpha^4 + \beta^4 = 13^2 - 2(\alpha\beta)^2 = 169 - 2(36) = 97$$

## Common pitfalls

- **Wrong:** Applying Vieta's formula directly to a non-monic polynomial without adjusting for the leading coefficient
  - Why it fails: The standard Vieta formulae are derived for monic polynomials, leading coefficient ≠ 1 changes all symmetric sum values
  - Correct: Always divide all terms of the polynomial by the leading coefficient to make it monic before applying the formula
- **Wrong:** Mixing up the sign of the product term for odd-degree polynomials
  - Why it fails: Signs alternate with the power of the root, and it is easy to forget the final sign for odd degree polynomials
  - Correct: Remember the general rule: product of roots = $(-1)^n \times$ constant term of the monic polynomial
- **Wrong:** Swapping the substitution when forming a new polynomial, substituting $x = 2y + 1$ instead of $y = 2x + 1$
  - Why it fails: Confusion between which variable represents the original root and which represents the new transformed root
  - Correct: Always define $y$ as the new root, set $y = f(x)$ then rearrange to get $x = f^{-1}(y)$ for substitution
- **Wrong:** Trying to calculate a non-symmetric function using only Vieta's formula
  - Why it fails: Non-symmetric functions change when roots are swapped, so they cannot be expressed from just symmetric sums of roots
  - Correct: Non-symmetric expressions require additional information, e.g. that one root is twice another, to solve

## Cheatsheet

| Polynomial (monic) | Vieta's Formulae | Common Identities |
| --- | --- | --- |
| $x^2 + ax + b = 0$<br>roots $\alpha, \beta$ | $\alpha+\beta = -a$<br>$\alpha\beta = b$ | $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$<br>$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}$ |
| $x^3 + ax^2 + bx + c = 0$<br>roots $\alpha, \beta, \gamma$ | $\sum\alpha = -a$<br>$\sum\alpha\beta = b$<br>$\alpha\beta\gamma = -c$ | $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$<br>$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\sum\alpha\beta}{\alpha\beta\gamma}$ |
| $x^4 + ax^3 + bx^2 + cx + d = 0$<br>roots $\alpha, \beta, \gamma, \delta$ | $\sum\alpha = -a$<br>$\sum\alpha\beta = b$<br>$\sum\alpha\beta\gamma = -c$<br>$\alpha\beta\gamma\delta = d$ | New polynomial with roots $y = kx + c$: substitute $x = \frac{y-c}{k}$ into original |

## What's next

Roots of polynomial equations and Vieta's formulae are foundational for almost all further work on polynomials in the CIE 9231 syllabus. This topic links directly to work on complex roots, the factor theorem, and solving higher degree polynomial equations, and also supports algebraic work in calculus and linear algebra. Mastery of symmetric sums will also help you with problems involving sums of series and coordinate geometry later in the course. The skills you learn here are also widely used in university-level mathematics.

- [Rational functions and graphs](https://www.owlsprep.com/study/cie-9231-u1-rational-functions-and-graphs/)
- [Summation of Series](https://www.owlsprep.com/study/cie-9231-u1-summation-of-series/)
- [Matrices](https://www.owlsprep.com/study/cie-9231-u1-matrices/)

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