Roots of polynomial equations · Further Mathematics 9231

Learning objectives

Write down the relationships between the coefficients of a quadratic, cubic or quartic equation and the elementary symmetric functions of its roots

Evaluate Σα², Σα³ and Σ1/α from the elementary symmetric functions using standard identities

Form a new polynomial equation whose roots are a given function of the original roots by substituting and eliminating x

Derive and use a recurrence relation for the power sums Sₙ = Σαⁿ, and solve problems in which an extra condition is imposed on the roots

Key definitions

Elementary symmetric functions

For roots α, β, γ these are Σα = α + β + γ, Σαβ = αβ + βγ + γα and αβγ. Any expression unchanged by permuting the roots can be written in terms of them.

Vieta relationships

The equalities linking the coefficients of a polynomial to the symmetric functions of its roots, for example Σα = −b/a for ax² + bx + c = 0.

Power sum

Sₙ = Σαⁿ = αⁿ + βⁿ + γⁿ + …, the sum of the nth powers of all the roots. S₀ equals the number of roots.

Substitution method

The technique of forming a new equation with roots y = f(α) by writing x = g(y) and substituting into the original equation, so that x is eliminated.

Recurrence for power sums

A relation such as Sₙ + pSₙ₋₁ + qSₙ₋₂ + rSₙ₋₃ = 0, obtained by multiplying the original equation by a power of x and summing over every root.

Explanation

The relationships between roots and coefficients. If α and β are the roots of ax² + bx + c = 0, then a(x − α)(x − β) = ax² + bx + c, and comparing coefficients gives Σα = α + β = −b/a and αβ = c/a. The same comparison for the cubic ax³ + bx² + cx + d = 0 gives Σα = −b/a, Σαβ = c/a and αβγ = −d/a, and for the quartic ax⁴ + bx³ + cx² + dx + e = 0 it gives Σα = −b/a, Σαβ = c/a, Σαβγ = −d/a and αβγδ = e/a. The signs alternate, and the divisor is always the leading coefficient a. Notice that these hold whether the roots are real or complex, so you never need to solve the equation to use them.

Symmetric functions built from the elementary ones. Squaring the sum of the roots gives (Σα)² = Σα² + 2Σαβ, so Σα² = (Σα)² − 2Σαβ. For a cubic the identity α³ + β³ + γ³ = (Σα)³ − 3ΣαΣαβ + 3αβγ gives Σα³ directly. For the reciprocals, Σ1/α = (βγ + γα + αβ)/(αβγ) = Σαβ/(αβγ), which for the cubic equals (c/a) ÷ (−d/a) = −c/d. In the same way Σ1/α² = (Σ1/α)² − 2Σ1/(αβ) and Σ(α + k) = Σα + 3k for a cubic. Write every quantity in terms of Σα, Σαβ and αβγ first, then substitute the numbers; that order keeps the algebra short and the signs under control.

Forming a new equation. To find the equation whose roots are y = f(α), rearrange to x = g(y) and substitute into the original equation, then clear fractions. For roots α + k put x = y − k; for roots kα put x = y/k; for roots 1/α put x = 1/y, which multiplied by yⁿ simply reverses the coefficients; for roots α² put x = √y, and since the square root is awkward, isolate the terms with odd powers on one side and square. For example, from x³ − 2x² + x − 5 = 0 the equation with roots 2α is obtained from x = y/2: y³/8 − y²/2 + y/2 − 5 = 0, and multiplying by 8 gives y³ − 4y² + 4y − 40 = 0. Always state the substitution you are using before you begin.

The power sum recurrence. Each root of x³ + px² + qx + r = 0 satisfies α³ + pα² + qα + r = 0. Multiplying by αⁿ⁻³ gives αⁿ + pαⁿ⁻¹ + qαⁿ⁻² + rαⁿ⁻³ = 0, and summing over all three roots gives Sₙ + pSₙ₋₁ + qSₙ₋₂ + rSₙ₋₃ = 0 for n ≥ 3. The starting values are S₀ = 3 (there are three roots, each contributing α⁰ = 1), S₁ = −p and S₂ = p² − 2q. The recurrence then produces S₃, S₄, S₅ and so on one line at a time, which is far quicker than expanding an identity. The same argument works for a quartic, where four consecutive power sums appear and S₀ = 4.

Extra conditions on the roots. When a question says the roots are in arithmetic progression, write them as a − d, a and a + d; the sum is then 3a, so the middle root is immediately one third of the sum of the roots, −p/3 for x³ + px² + qx + r = 0, and since it is a root it can be substituted back into the equation to give one equation in the unknown coefficients. If one root is the negative of another, say β = −α, then αβγ = −α²γ and Σα = γ, which quickly pins down γ. If the roots are in geometric progression, write them as a/k, a and ak, so the product is a³ and the middle root is the real cube root of the product. Choosing the symmetric labelling is the whole trick; after that the algebra is routine.

Formulae

Quadratic ax² + bx + c = 0: Σα = −b/a, αβ = c/a

Cubic ax³ + bx² + cx + d = 0: Σα = −b/a, Σαβ = c/a, αβγ = −d/a

Quartic ax⁴ + bx³ + cx² + dx + e = 0: Σα = −b/a, Σαβ = c/a, Σαβγ = −d/a, αβγδ = e/a

Σα² = (Σα)² − 2Σαβ

Σα³ = (Σα)³ − 3ΣαΣαβ + 3αβγ (cubic)

Σ1/α = Σαβ ÷ αβγ

Cubic x³ + px² + qx + r = 0: Sₙ + pSₙ₋₁ + qSₙ₋₂ + rSₙ₋₃ = 0, S₀ = 3

Worked examples

q: The cubic equation x³ − 4x² + 7x − 3 = 0 has roots α, β and γ. Find (a) Σα², (b) Σα³ using the power sum recurrence, and (c) the cubic equation with integer coefficients whose roots are 2α, 2β and 2γ.

steps: Read off the symmetric functions: Σα = −(−4)/1 = 4, Σαβ = 7/1 = 7 and αβγ = −(−3)/1 = 3.

(a) Σα² = (Σα)² − 2Σαβ = 4² − 2(7) = 16 − 14 = 2.

(b) Every root satisfies α³ = 4α² − 7α + 3. Summing over the three roots gives S₃ = 4S₂ − 7S₁ + 3S₀.

S₀ = 3, S₁ = Σα = 4 and S₂ = Σα² = 2, so S₃ = 4(2) − 7(4) + 3(3) = 8 − 28 + 9 = −11.

Multiply every term by 8: y³ − 8y² + 28y − 24 = 0. Check: the new sum of roots is 2Σα = 8 ✓, the new sum of pairs is 4Σαβ = 28 ✓ and the new product is 8αβγ = 24 ✓.

In context

Control engineers read the stability of a system straight from the coefficients of its characteristic polynomial, because those coefficients are the symmetric functions of the poles without the poles ever being computed.

A chemist balancing three coupled equilibrium constants often knows only their sum and product, which are exactly the symmetric functions of the unknown concentrations.

Common misconceptions

Students often forget the alternating signs and write αβγ = d/a for a cubic. In fact αβγ = −d/a: the sign alternates as the number of roots in each product increases.

Students often write Σα² = Σα × Σα. In fact (Σα)² also contains the cross terms, so Σα² = (Σα)² − 2Σαβ.

Students often substitute y = 2x into the equation instead of x = y/2. In fact the substitution must express the old variable in terms of the new one before it replaces x.

Students often take S₀ = 1 in the power sum recurrence. In fact S₀ = α⁰ + β⁰ + γ⁰ = 3 for a cubic, and equals the number of roots in general.

Study tips

Write the three (or four) symmetric functions out in a labelled line before doing anything else; most lost marks are sign slips in this first step.

When a new equation is wanted, state the substitution explicitly, for example "let y = α + 2, so x = y − 2", before substituting. Examiners award a method mark for it.

Check a new equation by testing one symmetric function: the sum of the new roots should match what the new coefficients predict.

If an equation for Sₙ is needed, derive the recurrence from the equation itself rather than quoting a formula, and state the range of n for which it is valid.

Summary

Coefficients give the symmetric functions with alternating signs and divisor a.

Σα² = (Σα)² − 2Σαβ; Σα³ = (Σα)³ − 3ΣαΣαβ + 3αβγ; Σ1/α = Σαβ ÷ αβγ.

New equations come from x = g(y): α + k needs x = y − k, kα needs x = y/k, 1/α reverses the coefficients.

Multiply the equation by a power of x and sum to get Sₙ + pSₙ₋₁ + qSₙ₋₂ + rSₙ₋₃ = 0 with S₀ = 3.

Roots in arithmetic progression are a − d, a, a + d, so the middle root is one third of the sum.

Check your understanding

Question 1: For 2x³ − 6x² + x − 4 = 0 the value of Σαβ is…

1. 3

2. −2

3. 0.5

Question 2: If Σα = 5 and Σαβ = 4 for a cubic, then Σα² equals…

1. 25

2. 17

3. 21

Question 3: To form the equation with roots α − 3 from an equation in x, substitute…

1. x = y + 3

2. x = y − 3

3. x = 3y