Maths › Further algebra and series › Roots of polynomials
Roots of polynomials
A polynomial's coefficients determine the symmetric functions of its roots. Their sum, their product and other symmetric combinations can all be found without solving the equation.
Builds on Polynomials and the factor theorem and Complex arithmetic and the Argand diagram.
IN THIS TOPIC
- Read the sum and product of roots straight from a polynomial's coefficients.
- Evaluate symmetric functions such as α² + β² + γ² without finding any root.
- Build a new polynomial whose roots are a transformation of the old ones.
- Solve a cubic or quartic with real coefficients when one complex root is given.
COMMON MISCONCEPTION
To find the sum of the roots of a cubic you must first solve the cubic.
Coefficients and roots
Write a cubic as a(z − α)(z − β)(z − γ) and expand. The z2 coefficient collects −a(α + β + γ), the z coefficient collects the pairwise products, and the constant collects −aαβγ. Comparing with az3 + bz2 + cz + d gives three relations:
The booklet prints no root relations of any degree, so these are yours to memorise. For a quadratic the first and last survive, so α + β = −b/a and αβ = c/a. For a quartic a fourth row appears, with the signs continuing to alternate. No solving happened anywhere.
WORKED EXAMPLE
Symmetric functions without solving
The cubic z³ + 2z² − 5z + 1 = 0 has roots α, β, γ. Find α² + β² + γ².
From the coefficients, α + β + γ = −2 and αβ + αγ + βγ = −5.
Square the sum: (α + β + γ)² = α² + β² + γ² + 2(αβ + αγ + βγ).
So α² + β² + γ² = (−2)² − 2(−5) = 14.
The roots themselves are a mess of surds. The symmetric combination never needed them.
New equations from old
Exam questions push one step further. Given an equation with roots α and β, find one whose roots are, say, 2α and 2β. The clean route is substitution. If w = 2z then z = w/2, and putting z = w/2 into the original equation produces a polynomial in w whose roots are exactly the doubled ones. You can also rebuild from sums and products, since the new sum and product follow from the old by arithmetic.
WORKED EXAMPLE
Squared roots from a quadratic
z² − 5z + 3 = 0 has roots α, β. Find a quadratic with roots α², β².
New sum: α² + β² = (α + β)² − 2αβ = 25 − 6 = 19.
New product: α²β² = (αβ)² = 9.
A quadratic is z² − (sum)z + (product), so z² − 19z + 9 = 0.
GUIDED PRACTICE
Doubled roots of a cubic
z³ − 6z² + 11z − 6 = 0 has roots α, β, γ. Find a cubic with roots 2α, 2β, 2γ.
Show the working
Substitute z = w/2: w³/8 − 6w²/4 + 11w/2 − 6 = 0.
Multiply through by 8: w³ − 12w² + 44w − 48 = 0.
Sense check with the known roots 1, 2, 3. The doubled set 2, 4, 6 has sum 12, pairwise sum 8 + 12 + 24 = 44 and product 48. Both routes agree.
One complex root gives you the rest
Complex roots of a polynomial with real coefficients arrive in conjugate pairs. Hand a candidate one complex root of a cubic and they have in fact been given two, so the third follows from the sum or the product with no factorising at all. The same trick reduces a quartic to a quadratic.
WORKED EXAMPLE
A cubic from one root
Given that 2 + 3i is a root of z³ − 5z² + 17z − 13 = 0, solve the equation.
The coefficients are real, so 2 − 3i is a root as well.
Sum of roots = 5, and (2 + 3i) + (2 − 3i) = 4, so the third root is 1.
The roots are 1, 2 + 3i, 2 − 3i. Check the product: 1 × (4 + 9) = 13, which matches −d/a.
The alternative route multiplies the conjugate pair into the real quadratic z² − 4z + 13 and divides. Both are fine, but the sum-of-roots line is quicker and much harder to get wrong under pressure. Watch the trap: the conjugate rule needs real coefficients, so it says nothing about z² − (2 + i)z + 5.
ASSESSMENT FOCUS
- Get the signs from the expansion, never from memory alone. The sum is −b/a and the signs alternate from there.
- Divide by the leading coefficient first when a ≠ 1. Forgetting it corrupts every relation.
- For α² + β² + γ², quote (Σα)² − 2Σαβ. The identity is expected, not derived from scratch.
- For transformed roots, state the substitution w = f(z) explicitly before rearranging.
- Given one complex root, write down its conjugate in the first line and say why you may.
CHECK YOURSELF
The equation 2z³ − 4z² + 3z − 7 = 0 has roots α, β, γ. Write down α + β + γ and αβγ.
Show a hint
Divide every coefficient by 2 before reading anything off.
Show the answer
Dividing by 2 gives z³ − 2z² + (3/2)z − 7/2 = 0, so α + β + γ = 2 and αβγ = 7/2. The pairwise sum, if asked, is 3/2.
Expand a(z − α)(z − β)…: root sums and products sit in the coefficients, signs alternating.
For transformed roots, substitute z in terms of w, or rebuild from the new sum and product.
Real coefficients force complex roots into conjugate pairs.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the roots of polynomials questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Read the sum and product of roots straight from a polynomial's coefficients.
- Evaluate symmetric functions such as α² + β² + γ² without finding any root.
- Build a new polynomial whose roots are a transformation of the old ones.
- Solve a cubic or quartic with real coefficients when one complex root is given.
Open the full revision checklist to see every objective in the course in one place.