D1 - Roots and coefficients | AQA A-Level Further Mathematics

You can find the sum of several unknown roots without solving the equation. Learn how a polynomial records sums and products of its roots, how to keep the signs correct, and how to recover sums of powers from that information.

Symmetric sums through quartic degree

The roots of $x^3-6x^2+11x-6=0$ are $\alpha,\beta,\gamma$ . Can you find $\alpha^2+\beta^2+\gamma^2$ without finding any individual root? The coefficients already contain the information we need.

Start with two known roots, $2$ and $5$ . Expanding their factors gives

(x-2)(x-5)=x^2-(2+5)x+2\times5=x^2-7x+10

The coefficient of $x$ is minus the sum of the roots; the constant is their product. For three roots, expanding one more bracket gives

(x-\alpha)(x-\beta)(x-\gamma) =x^3-(\alpha+\beta+\gamma)x^2 +(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma

Symmetric sums

A symmetric sum is unchanged when the roots are relabelled. Write $e_1$ for the sum of individual roots, $e_2$ for the sum of products of each distinct pair, and $e_3$ for the sum of products of each distinct triple. For four roots, $e_4$ is their product. Each pair or triple is counted once.

For a quartic, $e_2=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta$ has six terms, and $e_3=\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta$ has four. The pattern comes from choosing a root term from some brackets and $x$ from the others.

Roots and coefficients

In each row $a\ne0$ ; include every root with its multiplicity, real or complex.

Equation	$e_1$	$e_2$	$e_3$	$e_4$
$ax^2+bx+c=0$	$-b/a$	$c/a$	—	—
$ax^3+bx^2+cx+d=0$	$-b/a$	$c/a$	$-d/a$	—
$ax^4+bx^3+cx^2+dx+e=0$	$-b/a$	$c/a$	$-d/a$	$e/a$

These relationships are not in the formula booklet: learn them or derive them by expanding the factors.

Use the degree to decide what the last entry means: $e_2$ is the whole product only for a quadratic.

Worked example 1: reading a cubic

Question. Find $e_1,e_2,e_3$ for the roots of $x^3-6x^2+11x-6=0$ .

Plan. Compare the coefficients with $x^3-e_1x^2+e_2x-e_3$ .

\begin{aligned} -e_1&=-6 &&\text{compare the }x^2\text{ terms}\\ e_2&=11 &&\text{compare the }x\text{ terms}\\ -e_3&=-6 &&\text{compare constants} \end{aligned}

Answer. $e_1=6,\ e_2=11,\ e_3=6$ .

Check. Here $(x-1)(x-2)(x-3)$ expands to the given cubic. The roots give $1+2+3=6$ , $2+3+6=11$ and $1\times2\times3=6$ .

Now add one root, $4$ . This changes the degree and introduces a sum of triple products as well as the whole product.

Worked example 2: reading a quartic

Question. Find all four symmetric sums for $x^4-10x^3+35x^2-50x+24=0$ .

Plan. Match the coefficients to $x^4-e_1x^3+e_2x^2-e_3x+e_4$ .

\begin{aligned} e_1&=10,&e_2&=35 &&\text{first two coefficients}\\ e_3&=50,&e_4&=24 &&\text{last two coefficients} \end{aligned}

Answer. The sums are $10,35,50,24$ , respectively.

Check. Multiplying the previous cubic by $x-4$ gives this quartic. Its triple products sum to $6+8+12+24=50$ , and its product is $1\times2\times3\times4=24$ .

The relation works even when factorisation is inconvenient.

Watch out

Wrong for a quartic: “ $\alpha\beta\gamma\delta=-d/a$ ”. The coefficient $d$ belongs to $x$ , so $-d/a$ is the sum of four triple products. The whole product is $e/a$ .

Read a fresh quartic without solving it.

Signs, zero coefficients and repeated roots

Dividing the polynomial by its leading coefficient makes its first coefficient $1$ . This is the safest way to read signs, especially when the leading coefficient is negative.

The alternating signs belong to the factor expansion, not to a claim that the sums themselves are positive. A missing term has coefficient zero. It gives a zero symmetric sum; it need not make every root zero.

Multiplicity

A root has multiplicity $m$ when its factor occurs $m$ times. For $x(x-2)^2=0$ , the root list is $0,2,2$ , so the sum is $4$ , the pairwise sum is $4$ , and the product is $0$ . Repeated occurrences count separately.

An equation can be multiplied by any nonzero constant without changing its roots. Both the numerator coefficient and the leading coefficient change, so each ratio in the table stays the same.

Worked example 3: missing terms and multiplicity

Question. Find the four symmetric sums for $2x^4-8x^3+8x^2=0$ .

Plan. Insert the missing coefficients, then divide all coefficients by $2$ .

\begin{aligned} x^4-4x^3+4x^2+0x+0&=0 &&\text{normalise}\\ (e_1,e_2,e_3,e_4)&=(4,4,0,0) &&\text{compare alternating signs} \end{aligned}

Answer. $e_1=4,\ e_2=4,\ e_3=e_4=0$ .

Check. $2x^2(x-2)^2$ gives roots $0,0,2,2$ . Only the pair of nonzero occurrences contributes to $e_2$ , giving $2\times2=4$ ; every triple contains a zero.

Change only the overall multiplier, from $2$ to $-3$ . The roots and all their symmetric sums must stay unchanged.

Worked example 4: a negative leading coefficient

Question. Find the four symmetric sums for $-3x^4+12x^3-12x^2=0$ .

Plan. Divide by $-3$ , retaining each sign change.

\begin{aligned} x^4-4x^3+4x^2+0x+0&=0 &&\text{divide every term by }-3\\ e_1&=-12/(-3)=4\\ e_2&=(-12)/(-3)=4\\ e_3&=0,&e_4&=0 \end{aligned}

Answer. Again $(e_1,e_2,e_3,e_4)=(4,4,0,0)$ .

Check. The polynomial is $-3x^2(x-2)^2$ . Its roots, including multiplicity, are exactly the same as in Worked example 3.

A zero product guarantees at least one zero root, but a zero sum does not. For example, $x^2-1=0$ has roots $1,-1$ , whose sum is zero.

Watch out

Wrong: the sum for $-3x^4+12x^3-12x^2=0$ is $-12/3=-4$ . The leading coefficient is $-3$ , including its sign. Use $-12/(-3)=4$ .

Keep the zero terms and the negative leading coefficient visible in this check.

Identities and sums of powers

Squaring the sum of three roots produces their squares plus twice every pairwise product. Therefore the unknown roots can disappear from the calculation.

(\alpha+\beta+\gamma)^2 =\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma)

Write $p_k$ for the sum of powers of order $k$ of all roots, including multiplicities. Multiplying $e_1$ by $p_2$ , then subtracting $e_2p_1$ , cancels the mixed square terms; the triple products have been subtracted three times. This gives the second identity below.

Power sums from symmetric sums

For a cubic or quartic,

\begin{aligned} p_2&=e_1^2-2e_2\\ p_3&=e_1p_2-e_2e_1+3e_3 \end{aligned}

For a quartic, each root satisfies $r^4-e_1r^3+e_2r^2-e_3r+e_4=0$ . Adding this equation for all four root occurrences gives

p_4=e_1p_3-e_2p_2+e_3e_1-4e_4

The $p_2$ identity also holds for a quadratic; its $p_3$ identity has no $3e_3$ term. These identities are not in the formula booklet: derive them or learn them.

The coefficient of $e_4$ is $4$ because the constant is added once for each of the four root occurrences. No real-root assumption is needed.

Worked example 5: squares and cubes of cubic roots

Question. Without solving $x^3-6x^2+11x-6=0$ , find $p_2$ and $p_3$ .

Plan. Read $e_1,e_2,e_3$ , then substitute into the derived identities.

\begin{aligned} p_2&=6^2-2(11)=14 &&\text{remove pairwise terms}\\ p_3&=6(14)-11(6)+3(6)=36 &&\text{remove mixed cubes} \end{aligned}

Answer. $p_2=14$ and $p_3=36$ .

Check. The roots happen to be $1,2,3$ : $1+4+9=14$ and $1+8+27=36$ . They confirm the results but were not needed to obtain them.

Keep the same cubic and append its fourth root $4$ , as before. The new fourth-power calculation uses all four coefficient relations.

Worked example 6: powers of quartic roots

Question. Find $p_2,p_3,p_4$ for $x^4-10x^3+35x^2-50x+24=0$ , without solving it.

Plan. Use $e_1=10,e_2=35,e_3=50,e_4=24$ successively.

\begin{aligned} p_2&=100-70=30\\ p_3&=10(30)-35(10)+3(50)=100\\ p_4&=10(100)-35(30)+50(10)-4(24)\\ &=354 \end{aligned}

Answer. $p_2=30,\ p_3=100,\ p_4=354$ .

Check. The root list $1,2,3,4$ gives $1+16+81+256=354$ for the fourth powers; the square and cube totals exceed the previous totals by $16$ and $64$ , respectively.

This is the useful surprise: coefficients can answer a question about roots even when the roots cannot be expressed conveniently.

Watch out

Wrong: $\sum\alpha^2=(\sum\alpha)^2$ . The square on the right includes mixed products. Subtract $2e_2$ , so the opening problem gives $6^2-2(11)=14$ .

Use the same identities on a polynomial with different coefficients.

Quick Recap

Divide by the leading coefficient and compare with the alternating-sign factor expansion.
Pairwise and triple sums count each distinct selection of root occurrences once; repeated roots retain their multiplicities.
A quartic has four symmetric sums, ending with its product $e_4$ .
Power-sum identities eliminate individual roots; in particular, $p_2=e_1^2-2e_2$ .

The opening sum is $\alpha^2+\beta^2+\gamma^2=14$ , obtained directly from the coefficients without solving the cubic.

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D1 - Roots and coefficients

Symmetric sums through quartic degree

Signs, zero coefficients and repeated roots

Identities and sums of powers

Quick Recap

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