Roots of polynomial equations

CIE A-Level Further Mathematics· Further Pure 1 Unit 1: Polynomials· 25 min read

1. Vieta's Formulae for Polynomials★★☆☆☆⏱ 8 min

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.

📘 Definition

Vieta's Formulae (Monic Polynomials)

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

For a monic polynomial with roots $α_{1}, α_{2}, ..., α_{n}$ , Vieta's formulae state that the elementary symmetric sums of roots are related to coefficients with alternating signs.

\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 $2 x^{3} - 6 x^{2} + 4 x - 1 = 0$ has roots $α, β, γ$ . Find $α + β + γ$ , $α β + α γ + β γ$ and $α β γ$ .

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 $x^{3} + a x^{2} + b x + c = 0$ , so $a = - 3$ , $b = 2$ , $c = - \frac{1}{2}$
4
Apply Vieta's formula:
5
$\alpha+\beta+\gamma = -a = 3, \quad \alpha\beta + \alpha\gamma + \beta\gamma = b = 2, \quad \alpha\beta\gamma = -c = \frac{1}{2}$

Exam tip:

Always state you are using Vieta's formula in your working to earn method marks.

2. Calculating Symmetric Functions of Roots★★★☆☆⏱ 7 min

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.

📘 Definition

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 $2 x^{3} - 6 x^{2} + 4 x - 1 = 0$ from the previous example, find $α^{2} + β^{2} + γ^{2}$ and $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$ . We already know $α + β + γ = 3$ , $\sum α β = 2$ , $α β γ = \frac{1}{2}$ .

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
$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma} = \frac{2}{1/2} = 4$

3. Forming Polynomials with Transformed Roots★★★☆☆⏱ 6 min

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 $x^{3} - 4 x^{2} + 3 x - 2 = 0$ has roots $α, β, γ$ . Find a cubic equation with integer coefficients whose roots are $2 α + 1, 2 β + 1, 2 γ + 1$ .

1
Let $y$ be the new root, so $y = 2 x + 1$ 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
$y^3 - 19y^2 + 47y - 45 = 0$

Exam tip:

Always check your final polynomial has integer coefficients if the question asks for this.

4. Sum of Higher Powers of Roots★★★★☆⏱ 4 min

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 $x^{2} - 5 x + 6 = 0$ has roots $α$ and $β$ . Find $α^{4} + β^{4}$ .

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
$(\alpha^2 + \beta^2)^2 = \alpha^4 + 2\alpha^2\beta^2 \implies \alpha^4 + \beta^4 = 13^2 - 2(\alpha\beta)^2 = 169 - 2(36) = 97$

5. Common Pitfalls

Wrong move:

Applying Vieta's formula directly to a non-monic polynomial without adjusting for the leading coefficient

Why:

The standard Vieta formulae are derived for monic polynomials, leading coefficient ≠ 1 changes all symmetric sum values

Correct move:

Always divide all terms of the polynomial by the leading coefficient to make it monic before applying the formula

Wrong move:

Mixing up the sign of the product term for odd-degree polynomials

Why:

Signs alternate with the power of the root, and it is easy to forget the final sign for odd degree polynomials

Correct move:

Remember the general rule: product of roots = $(- 1)^{n} \times$ constant term of the monic polynomial

Wrong move:

Swapping the substitution when forming a new polynomial, substituting $x = 2 y + 1$ instead of $y = 2 x + 1$

Why:

Confusion between which variable represents the original root and which represents the new transformed root

Correct move:

Always define $y$ as the new root, set $y = f (x)$ then rearrange to get $x = f^{- 1} (y)$ for substitution

Wrong move:

Trying to calculate a non-symmetric function using only Vieta's formula

Why:

Non-symmetric functions change when roots are swapped, so they cannot be expressed from just symmetric sums of roots

Correct move:

Non-symmetric expressions require additional information, e.g. that one root is twice another, to solve

6. Quick Reference Cheatsheet

Polynomial (monic)	Vieta's Formulae	Common Identities
$x^{2} + a x + b = 0$ roots $α, β$	$α + β = - a$ $α β = b$	$α^{2} + β^{2} = (α + β)^{2} - 2 α β$ $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$
$x^{3} + a x^{2} + b x + c = 0$ roots $α, β, γ$	$\sum α = - a$ $\sum α β = b$ $α β γ = - c$	$\sum α^{2} = (\sum α)^{2} - 2 \sum α β$ $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{\sum α β}{α β γ}$
$x^{4} + a x^{3} + b x^{2} + c x + d = 0$ roots $α, β, γ, δ$	$\sum α = - a$ $\sum α β = b$ $\sum α β γ = - c$ $α β γ δ = d$	New polynomial with roots $y = k x + c$ : substitute $x = \frac{y - c}{k}$ into original

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Find transformed cubic polynomial
2023 · 1
Calculate symmetric sum of roots
2021 · 1
Use Vieta for quartic roots

Going deeper

syllabusCIE 9231 Further Maths Syllabus

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.