Maths › Further algebra and series › Roots of polynomials
Roots of polynomials
A polynomial's coefficients broadcast everything about its roots: their sum, their product, and every symmetric combination, all without solving anything.
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 linear transformation of the old ones.
WHAT YOU PROBABLY THINK
To find the sum of the roots of a cubic you must first solve the cubic.
Coefficients talk
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 facts for free:
For a quadratic the first and last survive: 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, which sinks the opening claim.
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, 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. Alternatively, rebuild from sums and products: 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.
YOUR TURN
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.
THE EXAM BIT
- Get the signs from the expansion, never from memory alone: sum is −b/a, product alternates 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.
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
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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.
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.
CHECK YOUR PROGRESS
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- 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 linear transformation of the old ones.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.