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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9231Paper 1 · Further Pure Mathematics 1Roots of polynomial equationsRoots of polynomial equations
Roots of polynomial equations · Subtopic 1 of 1

Roots of Polynomial Equations

For ax2+bx+c=0ax^2+bx+c=0: α+β=−ba\alpha+\beta=-\frac ba, αβ=ca\alpha\beta=\frac ca. Cubics and quartics have similar sums of roots taken one, two, three at a time. These let you evaluate expressions in the roots, and build new equations, without solving.

Cambridge 9231 Paper 1Edexcel YFM01 WFM01Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 40 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • State the sums and products of roots of quadratics, cubics and quartics.
  • Evaluate symmetric expressions such as α2+β2\alpha^2+\beta^2 or 1α+1β\frac1\alpha+\frac1\beta.
  • Form an equation whose roots are related to the roots of a given equation.
Pause and think

(x−2)(x−5)=x2−7x+10(x-2)(x-5)=x^2-7x+10. What do 77 and 1010 have to do with the roots 22 and 55?

Section 2 of 14

Core idea

Start hereAnswers without solving

Suppose someone asks for the sum of the squares of the three roots of x3+2x2−x+3=0x^3+2x^2-x+3=0. Solving that cubic is horrible. But the coefficients already "know" the sums and products of the roots, and that's enough to answer exactly.

The big idea

If a polynomial has roots α,β,…\alpha,\beta,\ldots, expanding a(x−α)(x−β)⋯a(x-\alpha)(x-\beta)\cdots shows the coefficients are (signed) sums of products of the roots. Express what you want in terms of those sums.

1. Where the relations come from

Quadratic
  1. ax2+bx+c=a(x−α)(x−β)ax^2+bx+c=a(x-\alpha)(x-\beta)
    Same roots, same leading coefficient.
  2. =ax2−a(α+β)x+aαβ=ax^2-a(\alpha+\beta)x+a\alpha\beta
    Expand.
  3. b=−a(α+β),c=aαβb=-a(\alpha+\beta),\quad c=a\alpha\beta
    Compare coefficients.
  4. α+β=−ba,αβ=ca\alpha+\beta=-\frac ba,\quad\alpha\beta=\frac ca
    Rearrange.
Cubic
  1. a(x−α)(x−β)(x−γ)a(x-\alpha)(x-\beta)(x-\gamma)
    Factorised form.
  2. =a[x3−(∑α)x2+(∑αβ)x−αβγ]=a\left[x^3-(\textstyle\sum\alpha)x^2+(\sum\alpha\beta)x-\alpha\beta\gamma\right]
    Expand: ∑αβ=αβ+βγ+γα\sum\alpha\beta=\alpha\beta+\beta\gamma+\gamma\alpha.
  3. ∑α=−ba, ∑αβ=ca, αβγ=−da\sum\alpha=-\frac ba,\ \sum\alpha\beta=\frac ca,\ \alpha\beta\gamma=-\frac da
    Compare with ax3+bx2+cx+dax^3+bx^2+cx+d.
Quartic: ∑α=−ba\sum\alpha=-\frac ba, ∑αβ=ca\sum\alpha\beta=\frac ca, ∑αβγ=−da\sum\alpha\beta\gamma=-\frac da, αβγδ=ea\alpha\beta\gamma\delta=\frac ea.
∑α=−ba,∑αβ=ca,∑αβγ=−da,αβγδ=ea\sum\alpha=-\frac ba,\quad\sum\alpha\beta=\frac ca,\quad\sum\alpha\beta\gamma=-\frac da,\quad\alpha\beta\gamma\delta=\frac ea
where
∑αβ\sum\alpha\beta
sum of products of pairs of roots.
a,b,c,d,ea,b,c,d,e
coefficients, highest power first.
Key takeawayAlternating signs Minus, plus, minus, plus: the sign pattern is the same for every degree.

2. Symmetric functions

∑α2\sum\alpha^2
  1. (α+β+γ)2=α2+β2+γ2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)
    Expand the square.
  2. ∑α2=(∑α)2−2∑αβ\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta
    Rearrange.
Worked example∑α2\sum\alpha^2 for a cubic

x3+2x2−x+3=0x^3+2x^2-x+3=0 has roots α,β,γ\alpha,\beta,\gamma. Find α2+β2+γ2\alpha^2+\beta^2+\gamma^2.

Goal
∑α2\sum\alpha^2.
Strategy
Relations, then the identity.
  1. ∑α=−2,∑αβ=−1\sum\alpha=-2,\quad\sum\alpha\beta=-1
    a=1a=1, b=2b=2, c=−1c=-1.
  2. ∑α2=(−2)2−2(−1)=6\sum\alpha^2=(-2)^2-2(-1)=6
    Identity.
Answer66.
✓ CheckA sum of squares of real numbers must be ≥0\ge0 (and 6>06>0) ✓; complex roots could even make it negative.
Worked exampleUsing the equation itself

x3−3x+1=0x^3-3x+1=0 has roots α,β,γ\alpha,\beta,\gamma. Find ∑α3\sum\alpha^3.

Goal
∑α3\sum\alpha^3.
Strategy
Each root satisfies the equation: α3=3α−1\alpha^3=3\alpha-1.
  1. ∑α3=3∑α−3\sum\alpha^3=3\sum\alpha-3
    Add the three equations.
  2. ∑α=0⇒∑α3=−3\sum\alpha=0\Rightarrow\sum\alpha^3=-3
    No x2x^2 term.
Answer−3-3.
✓ CheckIdentity ∑α3=(∑α)3−3∑α∑αβ+3αβγ=0−0+3(−1)=−3\sum\alpha^3=(\sum\alpha)^3-3\sum\alpha\sum\alpha\beta+3\alpha\beta\gamma=0-0+3(-1)=-3 ✓.

3. Transformed roots

Substitution method
  1. y=2x⇒x=y2y=2x\Rightarrow x=\tfrac y2
    New roots are 2α,2β2\alpha,2\beta: invert the transformation.
  2. (y2)2−5(y2)+3=0\left(\tfrac y2\right)^2-5\left(\tfrac y2\right)+3=0
    xx satisfies the old equation.
  3. y2−10y+12=0y^2-10y+12=0
    Multiply by 4.
Check: new sum 2(α+β)=102(\alpha+\beta)=10 ✓, new product 4αβ=124\alpha\beta=12 ✓.
Common misconceptionWrong direction For roots α+1\alpha+1, substitute x=y−1x=y-1, not y+1y+1.

Summary

  • Signs alternate: −ba,+ca,−da,+ea-\frac ba,+\frac ca,-\frac da,+\frac ea.
  • ∑α2=(∑α)2−2∑αβ\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta.
  • Transformed roots: substitute x=x= inverse.
Key words
Symmetric function
Expression unchanged when roots are swapped.
HistoryViète's formulas. François Viète, a French lawyer and codebreaker, wrote down the relations between roots and coefficients in the 1590s. They are still called Vieta's formulas.
Fun factNo formula for quintics. Quadratics, cubics and quartics have formulas for their roots, but Niels Abel proved in 1824 that general quintics do not. Root relations still work for every degree.
Section 3 of 14

Key formulae

Quadraticα+β=−ba,αβ=ca\alpha+\beta=-\frac ba,\qquad\alpha\beta=\frac ca
Cubic∑α=−ba,∑αβ=ca,αβγ=−da\sum\alpha=-\frac ba,\quad\sum\alpha\beta=\frac ca,\quad\alpha\beta\gamma=-\frac da
Quartic∑α=−ba, ∑αβ=ca, ∑αβγ=−da, αβγδ=ea\sum\alpha=-\frac ba,\ \sum\alpha\beta=\frac ca,\ \sum\alpha\beta\gamma=-\frac da,\ \alpha\beta\gamma\delta=\frac ea
Sum of squares∑α2=(∑α)2−2∑αβ\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta
TrickSigns alternate. Going along the coefficients, the sums of roots one, two, three, four at a time are −ba,+ca,−da,+ea-\frac ba,+\frac ca,-\frac da,+\frac ea: the signs alternate, starting with minus.
Section 4 of 14

Worked examples

Example 1 · Symmetric function0 / 2 steps
2x3−3x2+4x−5=02x^3-3x^2+4x-5=0 has roots α,β,γ\alpha,\beta,\gamma. Find α2+β2+γ2\alpha^2+\beta^2+\gamma^2.
    Example 2 · New equation0 / 2 steps
    x2−5x+3=0x^2-5x+3=0 has roots α,β\alpha,\beta. Find an equation with roots 2α,2β2\alpha,2\beta.
      Section 5 of 14

      Explore the graph

      Roots and coefficients
      Loading the interactive graph…
      Try this
      1. The yy-intercept is pqpq, the product of the roots. Check as you move the sliders.
      2. Where is the vertex? Its xx-coordinate is the average of the roots, p+q2\frac{p+q}2.

      y=x2−(p+q)x+pqy=x^2-(p+q)x+pq: the coefficients are the sum and product of the roots.

      In the real worldControl engineering. Engineers check whether a system is stable from the coefficients of its characteristic polynomial, using root relations, without ever solving for the roots.
      Section 6 of 14

      Common mistakes

      Incorrectα+β=ba\alpha+\beta=\frac ba
      Correct−ba-\frac ba
      Expanding a(x−α)(x−β)a(x-\alpha)(x-\beta) gives −a(α+β)x-a(\alpha+\beta)x.
      Incorrectα2+β2=(α+β)2\alpha^2+\beta^2=(\alpha+\beta)^2
      Correct(α+β)2−2αβ(\alpha+\beta)^2-2\alpha\beta
      Squaring a sum gives cross terms.
      IncorrectForgetting aa when a≠1a\ne1
      CorrectDivide every coefficient by aa
      The relations use ba\frac ba, ca\frac ca, …

      Me: solve the cubic, then square the roots

      Root relations: (∑α)2−2∑αβ(\sum\alpha)^2-2\sum\alpha\beta. Done. 😎

      Section 7 of 14

      Exam tip

      Exam tipState the relations first. Write ∑α=…\sum\alpha=\ldots, ∑αβ=…\sum\alpha\beta=\ldots, αβγ=…\alpha\beta\gamma=\ldots on the first line. They usually carry a B mark and you will need them all.
      Section 8 of 14

      Guided practice

      SA-FP-ROOT-G012 marksL2 StandardNon-calculatorSymmetric functions
      x2−5x+3=0x^2-5x+3=0 has roots α\alpha and β\beta. Find α2+β2\alpha^2+\beta^2.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-ROOT-0012 marksL2 StandardNon-calculatorSymmetric functions
      For the same equation x2−5x+3=0x^2-5x+3=0, find 1α+1β\dfrac1\alpha+\dfrac1\beta.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-ROOT-0021 markL1 FoundationNon-calculatorRoot relations
      2x3−3x2+4x−5=02x^3-3x^2+4x-5=0 has roots α,β,γ\alpha,\beta,\gamma. Find αβγ\alpha\beta\gamma.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-ROOT-0033 marksL2 StandardNon-calculatorSymmetric functions
      x3+2x2−x+3=0x^3+2x^2-x+3=0 has roots α,β,γ\alpha,\beta,\gamma. Find α2+β2+γ2\alpha^2+\beta^2+\gamma^2.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-ROOT-0043 marksL2 StandardNon-calculatorTransformed roots
      x2−5x+3=0x^2-5x+3=0 has roots α,β\alpha,\beta. The equation y2+py+q=0y^2+py+q=0 has roots 2α2\alpha and 2β2\beta. Find qq.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-ROOT-0054 marksL4 ChallengingNon-calculatorSymmetric functions
      x3−3x+1=0x^3-3x+1=0 has roots α,β,γ\alpha,\beta,\gamma. Find α3+β3+γ3\alpha^3+\beta^3+\gamma^3.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Signs alternate: −ba,+ca,−da,+ea-\frac ba,+\frac ca,-\frac da,+\frac ea.
      • ∑α2=(∑α)2−2∑αβ\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta.
      • New roots: substitute the inverse of the transformation.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-ROOT-Q011 markL1 FoundationNon-calculatorRoot relations
      The product of the roots of ax2+bx+c=0ax^2+bx+c=0 is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      9231 AS & A Level Further Mathematics
      Where
      Paper 1 · 1.1 Roots of polynomial equations
      Expected
      Relations between roots and coefficients for quadratics, cubics and quartics; symmetric functions; substitution to form new equations.
      Calculator
      Permitted on all papers; exact answers are expected where the question asks.
      Typical style
      Typical opener: "Find the value of α² + β² + γ²", followed by "find an equation whose roots are …".
      Open in
      9231 · Paper 1
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).