Home / A-Level Further Maths 9231 / Roots of polynomial equations / Sums and products of roots
Cambridge International AS & A Level · Topic 1.1
A-Level Further Maths 9231 / Topic 1.1 / AS & A Level

Read the roots without solving.

Every polynomial carries information about its roots in its coefficients. Expand \(a(x - \alpha)(x - \beta)(x - \gamma)\) and the sum of the roots, the sum of their products in pairs and their product all appear, each as a simple fraction of the coefficients. You can use these to form equations from their roots and to find unknown coefficients.

Topic 1.1All roots of polynomial equations lessons →
The key idea

For \(ax^{3} + bx^{2} + cx + d = 0\) with roots \(\alpha, \beta, \gamma\): \(\sum \alpha = -\dfrac{b}{a}\), \(\sum \alpha\beta = \dfrac{c}{a}\), \(\alpha\beta\gamma = -\dfrac{d}{a}\). The signs alternate, starting with minus. A quadratic stops after two relations; a quartic adds \(\sum \alpha\beta\gamma = -\dfrac{d}{a}\) and \(\alpha\beta\gamma\delta = \dfrac{e}{a}\).

x y −3 1 2 6 y = x³ − 7x + 6 roots −3, 1, 2 Σα = 0 no x² term Σαβ = −7 x coefficient αβγ = −6 minus the constant
Fig. 1: the cubic y = x³ - 7x + 6 crosses the x-axis at -3, 1 and 2. Their sum, their pair products and their product can all be read off the coefficients without solving
Section 01

Expand the factors, match the coefficients.

If a quadratic \(ax^{2} + bx + c = 0\) has roots \(\alpha\) and \(\beta\), then \(ax^{2} + bx + c \equiv a(x - \alpha)(x - \beta) = a\left(x^{2} - (\alpha + \beta)x + \alpha\beta\right)\). Comparing coefficients gives \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\). The same idea works for a cubic with roots \(\alpha, \beta, \gamma\).

\[a(x - \alpha)(x - \beta)(x - \gamma) = a\left(x^{3} - (\alpha + \beta + \gamma)x^{2} + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma\right)\]

Compare with \(ax^{3} + bx^{2} + cx + d\). We write \(\sum \alpha\) for the sum of the roots and \(\sum \alpha\beta\) for the sum of the products of the roots taken two at a time.

EquationSum of rootsPairsProduct
\(ax^{2} + bx + c = 0\)\(\alpha + \beta = -\frac{b}{a}\)\(\alpha\beta = \frac{c}{a}\)
\(ax^{3} + bx^{2} + cx + d = 0\)\(\sum \alpha = -\frac{b}{a}\)\(\sum \alpha\beta = \frac{c}{a}\)\(\alpha\beta\gamma = -\frac{d}{a}\)
Worked example Read the three relations
The roots of \(2x^{3} - 5x^{2} + 3x + 7 = 0\) are \(\alpha, \beta, \gamma\). Write down \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\).
  1. \(a = 2,\ b = -5,\ c = 3,\ d = 7\).
  2. \(\sum \alpha = -\dfrac{b}{a} = -\dfrac{-5}{2} = \dfrac{5}{2}\).
  3. \(\sum \alpha\beta = \dfrac{c}{a} = \dfrac{3}{2}\).
  4. \(\alpha\beta\gamma = -\dfrac{d}{a} = -\dfrac{7}{2}\).
Answer: \(\sum \alpha = \frac{5}{2},\ \sum \alpha\beta = \frac{3}{2},\ \alpha\beta\gamma = -\frac{7}{2}\)
Worked example A missing term is a zero coefficient
The cubic \(x^{3} - 7x + 6 = 0\) has roots \(-3\), \(1\) and \(2\) (Fig. 1). Find \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\) from the coefficients, and check them against the roots.
  1. Write it in full: \(x^{3} + 0x^{2} - 7x + 6 = 0\), so \(a = 1,\ b = 0,\ c = -7,\ d = 6\).
  2. From the coefficients: \(\sum \alpha = 0\), \(\sum \alpha\beta = -7\), \(\alpha\beta\gamma = -6\).
  3. Check: \(-3 + 1 + 2 = 0\); \((-3)(1) + (1)(2) + (2)(-3) = -3 + 2 - 6 = -7\); \((-3)(1)(2) = -6\).
Answer: \(\sum \alpha = 0,\ \sum \alpha\beta = -7,\ \alpha\beta\gamma = -6\), all confirmed
Your turn
The roots of \(3x^{2} - 4x - 2 = 0\) are \(\alpha\) and \(\beta\). Use \(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\) to find \(\alpha^{2} + \beta^{2}\) without solving.
Examiner trap

The sum of the roots is \(-\dfrac{b}{a}\), not \(\dfrac{b}{a}\), and for a cubic the product is \(-\dfrac{d}{a}\). Divide by \(a\) every time: forgetting it is only safe when \(a = 1\).

Section 02

Quartics, and building an equation from its roots.

A quartic \(ax^{4} + bx^{3} + cx^{2} + dx + e = 0\) with roots \(\alpha, \beta, \gamma, \delta\) has four relations. \(\sum \alpha\beta\) now has six terms (every pair) and \(\sum \alpha\beta\gamma\) has four (every triple). Run the process backwards and the relations build the equation: the signs in front of the sums alternate.

Worked example Four relations from a quartic
The roots of \(2x^{4} + 3x^{3} - x^{2} + 5x - 4 = 0\) are \(\alpha, \beta, \gamma, \delta\). Write down \(\sum \alpha\), \(\sum \alpha\beta\), \(\sum \alpha\beta\gamma\) and \(\alpha\beta\gamma\delta\).
  1. \(a = 2,\ b = 3,\ c = -1,\ d = 5,\ e = -4\).
  2. \(\sum \alpha = -\dfrac{3}{2}\) and \(\sum \alpha\beta = -\dfrac{1}{2}\).
  3. \(\sum \alpha\beta\gamma = -\dfrac{5}{2}\) and \(\alpha\beta\gamma\delta = \dfrac{-4}{2} = -2\).
Answer: \(-\frac{3}{2},\ -\frac{1}{2},\ -\frac{5}{2},\ -2\)
Worked example Form a cubic with integer coefficients
The roots \(\alpha, \beta, \gamma\) of a cubic satisfy \(\sum \alpha = \dfrac{1}{2}\), \(\sum \alpha\beta = -2\) and \(\alpha\beta\gamma = \dfrac{3}{2}\). Find the cubic equation with integer coefficients.
  1. \(x^{3} - (\sum \alpha)x^{2} + (\sum \alpha\beta)x - \alpha\beta\gamma = 0\).
  2. \(x^{3} - \dfrac{1}{2}x^{2} - 2x - \dfrac{3}{2} = 0\).
  3. Multiply by 2 to clear the fractions.
Answer: \(2x^{3} - x^{2} - 4x - 3 = 0\)
Your turn
Find the quartic equation with roots \(1, -1, 2, 3\) by working out the four sums.
Examiner trap

When you build an equation, the signs alternate: minus the sum, plus the pairs, minus the triples, plus the product of four. Copying the sums straight in as coefficients (all plus) is the most common error.

Stage 1 · Learn

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The roots of \(3x^{3} - 6x^{2} + x + 9 = 0\) are \(\alpha, \beta, \gamma\). Find \(\alpha + \beta + \gamma\).

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The roots of \(x^{4} - 2x^{3} + 5x - 1 = 0\) are \(\alpha, \beta, \gamma, \delta\). Find \(\sum \alpha\beta\).

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Which cubic equation has roots \(2, -1, 3\)?

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The roots of \(x^{2} - 5x + 3 = 0\) are \(\alpha\) and \(\beta\). Find \(\alpha^{2} + \beta^{2}\).

Section 03

Unknown coefficients and special roots.

Many questions describe the roots instead of giving them: the roots are in arithmetic progression, one root is double another, two roots have a given product, or one root is complex. Choose letters that build in the description, then use the relations. For an equation with REAL coefficients, complex roots come in conjugate pairs.

Worked example Roots in arithmetic progression
The roots of \(x^{3} - 6x^{2} + kx - 6 = 0\) are in arithmetic progression. Find \(k\) and the roots.
  1. Let the roots be \(a - d,\ a,\ a + d\). Then \(\sum \alpha = 3a = 6\), so \(a = 2\).
  2. \(x = 2\) is a root: \(8 - 24 + 2k - 6 = 0\), so \(2k = 22\) and \(k = 11\).
  3. Product: \((2 - d)(2)(2 + d) = 6\), so \(4 - d^{2} = 3\) and \(d = \pm 1\).
  4. Either sign gives the same set of roots.
Answer: \(k = 11\); the roots are \(1, 2, 3\)
Worked example One root double another
The equation \(x^{3} - 7x^{2} + 14x - k = 0\) has one root that is twice another. Find \(k\) and the roots.
  1. Let the roots be \(\alpha,\ 2\alpha,\ \beta\). Sum: \(3\alpha + \beta = 7\), so \(\beta = 7 - 3\alpha\).
  2. Pairs: \(2\alpha^{2} + \alpha\beta + 2\alpha\beta = 2\alpha^{2} + 3\alpha\beta = 14\).
  3. Substitute: \(2\alpha^{2} + 3\alpha(7 - 3\alpha) = 14 \Rightarrow -7\alpha^{2} + 21\alpha - 14 = 0 \Rightarrow \alpha^{2} - 3\alpha + 2 = 0\).
  4. \(\alpha = 1\) gives roots \(1, 2, 4\); \(\alpha = 2\) gives \(2, 4, 1\), the same set.
  5. Product: \(\alpha\beta\gamma = 8 = -\dfrac{-k}{1} = k\).
Answer: \(k = 8\); the roots are \(1, 2, 4\)
Your turn
Given that \(1 + 2i\) is a root of \(x^{3} + px^{2} + qx + 10 = 0\), where \(p\) and \(q\) are real, find the real root, \(p\) and \(q\).
Examiner trap

With three roots in arithmetic progression, do not write \(a,\ a + d,\ a + 2d\): the sum becomes \(3a + 3d\) and you have two unknowns where \(a - d,\ a,\ a + d\) gives \(a\) at once. And keep the sign in \(\alpha\beta\gamma = -\dfrac{d}{a}\) when the product is the relation you use.

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The roots of \(x^{3} - 9x^{2} + kx - 15 = 0\) are in arithmetic progression. Find \(k\).

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Given that \(1 - 3i\) is a root of \(x^{3} + ax^{2} + bx - 30 = 0\), where \(a\) and \(b\) are real, find \(a\) and \(b\).

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The roots \(\alpha, \beta, \gamma, \delta\) of a quartic satisfy \(\sum \alpha = 3\), \(\sum \alpha\beta = -2\), \(\sum \alpha\beta\gamma = 4\) and \(\alpha\beta\gamma\delta = -1\). Which is the quartic?

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Two of the roots of \(x^{3} - 4x^{2} + px + 12 = 0\) have product 6. Find \(p\).

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Six original Cambridge-style questions
Reading the relations from a cubic, \(\alpha^{2} + \beta^{2}\) for a quadratic, forming a cubic from its roots, roots in arithmetic progression, a quartic with a complex root and a root that is double another.