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How do the roots of a polynomial relate to its coefficients, and how do you sum series and split fractions?

Roots of polynomials and their relationships to coefficients, summation of series using standard results, the method of differences, partial fractions and the Maclaurin series.

A focused answer to the AQA A-Level Further Mathematics further algebra content, covering relationships between roots and coefficients, summation of series with standard results, the method of differences, partial fractions and the Maclaurin series.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Roots and coefficients
  3. Summing series
  4. The method of differences
  5. Maclaurin series

What this dot point is asking

AQA wants you to relate the roots of a polynomial to its coefficients, evaluate finite series using the standard results for sums of powers, use the method of differences for telescoping sums, decompose rational functions into partial fractions, and derive and use the Maclaurin series of a function.

Roots and coefficients

For a quadratic with roots α\alpha and β\beta, α+β=ba\alpha + \beta = -\frac{b}{a} and αβ=ca\alpha\beta = \frac{c}{a}. For a cubic ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0 with roots α,β,γ\alpha, \beta, \gamma, the sum of roots is ba-\frac{b}{a}, the sum of products in pairs αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alpha is ca\frac{c}{a}, and the product αβγ\alpha\beta\gamma is da-\frac{d}{a}. For a quartic the pattern continues: the symmetric functions alternate in sign as ba,+ca,da,+ea-\frac{b}{a}, +\frac{c}{a}, -\frac{d}{a}, +\frac{e}{a}. These are the elementary symmetric functions, and the standard exam tasks build derived quantities from them. The square-of-sum identity α2=(α)22αβ\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta and the cube identity α3=(α)33(α)(αβ)+3αβγ\sum\alpha^3 = (\sum\alpha)^3 - 3(\sum\alpha)(\sum\alpha\beta) + 3\alpha\beta\gamma recur often, as does forming a new equation whose roots are a transformation of the originals (such as α+1\alpha + 1, 2α2\alpha or 1α\frac{1}{\alpha}) by working out the new symmetric functions.

Summing series

The standard results are r=1nr=12n(n+1)\sum_{r=1}^{n} r = \frac{1}{2}n(n+1), r=1nr2=16n(n+1)(2n+1)\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1) and r=1nr3=14n2(n+1)2\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2. Split a polynomial summand into these pieces.

The method of differences

If a summand can be written as f(r)f(r+1)f(r) - f(r+1), the sum telescopes and almost all terms cancel, leaving only the first and last. Partial fractions are often the way to get this form.

Maclaurin series

The Maclaurin series of f(x)f(x) is f(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3+f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots, valid where the series converges. Standard expansions for exe^x, sinx\sin x, cosx\cos x and ln(1+x)\ln(1+x) are worth knowing.

When a summand is not a simple power of rr, look for the method of differences instead: a partial fraction split usually turns the term into f(r)f(r+1)f(r) - f(r+1) so the sum telescopes, leaving only the boundary terms.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20195 marksThe cubic equation x36x2+11x6=0x^3 - 6x^2 + 11x - 6 = 0 has roots α\alpha, β\beta and γ\gamma. Without solving the equation, find the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.
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Use the relationships between roots and coefficients for a cubic x3+px2+qx+r=0x^3 + px^2 + qx + r = 0.

Here α=6\sum\alpha = 6, αβ=11\sum\alpha\beta = 11 and αβγ=6\alpha\beta\gamma = 6.

The key identity is α2+β2+γ2=(α+β+γ)22(αβ+βγ+γα)\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha).

Substitute: α2+β2+γ2=622(11)=3622=14\alpha^2 + \beta^2 + \gamma^2 = 6^2 - 2(11) = 36 - 22 = 14.

Markers reward correct extraction of the symmetric functions and use of the squared-sum identity rather than solving for the roots.

AQA 20216 marksUse the method of differences to find r=1n1(2r1)(2r+1)\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)}, giving your answer as a single fraction in terms of nn.
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Split into partial fractions: 1(2r1)(2r+1)=12(12r112r+1)\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right).

So the sum is 12r=1n(12r112r+1)\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right).

Writing the terms out, 12[(113)+(1315)++(12n112n+1)]\frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right], the interior terms telescope.

Only the first and last survive: 12(112n+1)=122n2n+1=n2n+1\frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \frac{1}{2}\cdot\frac{2n}{2n+1} = \frac{n}{2n+1}.

Markers reward the partial fraction split (with the factor of 12\frac{1}{2}), the telescoping, and the simplified single fraction.

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