Paper 1 · Further Pure 1 ✦

Roots of Polynomial Equations

The link between a polynomial's coefficients and its roots — Vieta's relations, symmetric functions, and building new equations without ever solving the original.

Syllabus 1.1AlgebraDegrees 2, 3 & 4

What the examiner expects you to do

Recall & use the relations between roots and coefficients for equations of degree 2, 3 and 4.
Evaluate symmetric functions of the roots (e.g. $\sum α^{2}$ , $\sum \frac{1}{α}$ ).
Use a substitution to form an equation whose roots are simply related to the original's.
Solve problems with unknown coefficients using these relations.

§1 Relations between roots and coefficients

Every polynomial factorises over its roots. Multiplying that factored form out and comparing coefficients gives Vieta's relations — the roots' symmetric sums read straight off the coefficients.

◆ Vieta's relations

Quadratic $a x^{2} + b x + c = 0$ with roots $α, β$ :

α + β = - \frac{b}{a}, α β = \frac{c}{a} .

Cubic $a x^{3} + b x^{2} + c x + d = 0$ with roots $α, β, γ$ :

\sum α = - \frac{b}{a}, \sum α β = \frac{c}{a}, α β γ = - \frac{d}{a} .

Quartic $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ with roots $α, β, γ, δ$ :

\sum α = - \frac{b}{a}, \sum α β = \frac{c}{a}, \sum α β γ = - \frac{d}{a}, α β γ δ = \frac{e}{a} .

The signs alternate, and each sum runs over all groupings of that size.

Where it comes from (cubic)

If $α, β, γ$ are the roots, then $a x^{3} + b x^{2} + c x + d = a (x - α) (x - β) (x - γ)$ . Expand the right-hand side:

a [x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ] .

Compare coefficients with $a x^{3} + b x^{2} + c x + d$ : $b = - a \sum α$ , $c = a \sum α β$ , $d = - a α β γ$ — which rearrange to the relations above. $■$

★ A* tip

Always divide through by the leading coefficient $a$ first (or keep the $/ a$ ). Forgetting it is the single most common slip here.

§2 Symmetric functions of the roots

A symmetric function is unchanged if you swap the roots around — e.g. $\sum α^{2}$ , $\sum \frac{1}{α}$ , $(α - β)^{2}$ . Every one can be written using the basic sums above, so you never need the roots themselves.

◇ The workhorse identities

\sum α^{2} = (\sum α)^{2} - 2 \sum α β, (α - β)^{2} = (α + β)^{2} - 4 α β .

Sums of reciprocals only need a common denominator — the numerator is the next-to-last symmetric sum, the denominator is the product of all the roots — so the exact form depends on the degree:

quadratic: \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}, cubic: \sum \frac{1}{α} = \frac{\sum α β}{α β γ} .

✎ Worked example — a quadratic

$2 x^{2} - 5 x + 1 = 0$ has roots $α, β$ . Then $α + β = \frac{5}{2}$ and $α β = \frac{1}{2}$ . So

α^{2} + β^{2} = (\frac{5}{2})^{2} - 2 \cdot \frac{1}{2} = \frac{21}{4}, \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = 5, (α - β)^{2} = \frac{25}{4} - 2 = \frac{17}{4} .

✎ Worked example — a cubic

$x^{3} + 2 x^{2} - 5 x + 1 = 0$ : $\sum α = - 2, \sum α β = - 5, α β γ = - 1$ . Then

\sum α^{2} = (- 2)^{2} - 2 (- 5) = 14, \sum \frac{1}{α} = \frac{\sum α β}{α β γ} = \frac{- 5}{- 1} = 5 .

★ A* method — higher powers: substitute & add

For $\sum α^{3}$ and beyond, never expand. Use the fact that every root satisfies the equation: if $α, β, γ$ are the roots of $x^{3} + b x^{2} + c x + d = 0$ then $α^{3} + b α^{2} + c α + d = 0$ , and likewise for $β$ and $γ$ . Add the three copies:

\sum α^{3} + b \sum α^{2} + c \sum α + 3 d = 0 ⟹ \sum α^{3} = - b \sum α^{2} - c \sum α - 3 d .

Check it on $x^{3} + 2 x^{2} - 5 x + 1 = 0$ from above, where $\sum α = - 2$ and $\sum α^{2} = 14$ :

\sum α^{3} = - 2 (14) + 5 (- 2) - 3 (1) = - 41 .

Multiply by $α^{k}$ before adding to climb to $\sum α^{4}, \sum α^{5}, \dots$ ; divide by $α$ instead to drop to negative powers. The $3 d$ is the classic slip — you add three copies of the constant, not one.

§3 Forming a new equation

To find the equation whose roots are $y = g (α)$ , substitute $α = g^{- 1} (y)$ into the original and tidy up. No need to know the actual roots.

✎ Worked example — scaled roots

The roots of $x^{3} - 3 x + 1 = 0$ are $α, β, γ$ . Find the equation with roots $2 α, 2 β, 2 γ$ .

New root $y = 2 x \Rightarrow x = \frac{y}{2}$ . Substitute:

(\frac{y}{2})^{3} - 3 (\frac{y}{2}) + 1 = 0 \Rightarrow \frac{y^{3}}{8} - \frac{3 y}{2} + 1 = 0 \Rightarrow y^{3} - 12 y + 8 = 0 .

✎ Worked example — reciprocal roots

The roots of $3 x^{3} - x^{2} + 2 x - 5 = 0$ are $α, β, γ$ . Find the equation with roots $\frac{1}{α}, \frac{1}{β}, \frac{1}{γ}$ .

Put $y = \frac{1}{x} \Rightarrow x = \frac{1}{y}$ , then multiply through by $y^{3}$ :

\frac{3}{y^{3}} - \frac{1}{y^{2}} + \frac{2}{y} - 5 = 0 \Rightarrow 3 - y + 2 y^{2} - 5 y^{3} = 0 \Rightarrow 5 y^{3} - 2 y^{2} + y - 3 = 0 .

⚠ Trap — squared roots

For roots $y = α^{2}$ , you can't just write $x = \sqrt{y}$ and stop — isolate the surd and square to remove it, or build the new equation from its symmetric functions instead.

§4 Interactive: roots build the coefficients

Drag the three roots of a cubic $y = (x - r_{1}) (x - r_{2}) (x - r_{3})$ and watch the expanded coefficients — and Vieta's sums — update live. This is §1, in motion.

Root

r_{1}

-2.0

Root

r_{2}

0.5

Root

r_{3}

2.5

drag the roots ✎

The x-intercepts are the roots; the coefficients are their symmetric sums.

§5 Examiner traps & A* checklist

⚠ The four most common mark-losers

Dropping the leading coefficient — the relations use $- b / a$ , not $- b$ .
Sign slips in the alternating pattern ( $-, +, -, + \dots$ ).
Reading $\sum α β$ as $(\sum α) (\sum β)$ — it's the sum of products in pairs.
Giving a new equation of the wrong degree — a cubic's substitution yields a cubic.

§6 Video explainers (curated)

▶Roots ↔ coefficients (Vieta)
Curated search — degree 2/3/4 relations, worked.↗ ▶Symmetric functions of the roots
Curated search — Σα², Σ1/α, Σα³ by substitute & add.↗ ▶Forming a new equation
Curated search — substitution method, scaled & reciprocal roots.↗

§7 Check yourself

Score 0 / 5

Q1 · quadratic

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

Solution

$α + β = 6, α β = 4$ , so $α^{2} + β^{2} = 6^{2} - 2 (4) = 28$ .

Q2 · cubic · Vieta

For $x^{3} - 2 x^{2} + 3 x - 4 = 0$ with roots $α, β, γ$ , the value of $α + β + γ$ is:

Solution

$\sum α = - \frac{b}{a} = - \frac{- 2}{1} = 2$ .

Q3 · cubic · pairs

For the same cubic $x^{3} - 2 x^{2} + 3 x - 4 = 0$ , the value of $\sum α β$ is:

Solution

$\sum α β = \frac{c}{a} = \frac{3}{1} = 3$ (sum of products in pairs).

Q4 · new equation

$x^{2} - 3 x + 2 = 0$ has roots $α, β$ . The equation with roots $2 α, 2 β$ is:

Solution

Put $x = \frac{y}{2}$ : $\frac{y^{2}}{4} - \frac{3 y}{2} + 2 = 0$ , then $\times 4$ gives $y^{2} - 6 y + 8 = 0$ .

Q5 · in terms of a parameter

The roots of $x^{3} + p x + q = 0$ are $α, β, γ$ . Show that $α^{2} + β^{2} + γ^{2} = - 2 p$ .

Solution

There is no $x^{2}$ term, so $\sum α = 0$ ; and $\sum α β = p$ . Hence

\sum α^{2} = (\sum α)^{2} - 2 \sum α β = 0 - 2 p = - 2 p . ■

FP1 · Topic 1 · Roots of Polynomial Equations