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Edexcel A-level Further Maths revision notes

Further algebra and functions

Section CP-4
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
6 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section CP-4

Checked against Edexcel 9FM0 section CP-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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CP-4.1

Understand and use the relationship between roots and coefficients of polynomial equations up to quartic equations.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a monic cubic x3+ax2+bx+c=0x^3+ax^2+bx+c=0 with roots α,β,γ\alpha,\beta,\gamma, comparison with (xα)(xβ)(xγ)(x-\alpha)(x-\beta)(x-\gamma) gives sum of roots a-a, sum of pair products bb and product c-c.
  • A monic quartic follows the alternating pattern: the elementary symmetric sums are a,b,c,d-a,b,-c,d.
  • Divide a non-monic equation by its leading coefficient first.
  • Further expressions are built from these sums: squares use (α)2(\sum\alpha)^2, reciprocals use the ratio of the next-to-last symmetric sum to the product, and shifted products can be evaluated by substituting into the polynomial.
  • The roots need not be found explicitly.
Worked example

The roots of x36x2+5x2=0x^3-6x^2+5x-2=0 are α,β,γ\alpha,\beta,\gamma. Find α2+β2+γ2\alpha^2+\beta^2+\gamma^2.

  1. 1.α+β+γ=6\alpha+\beta+\gamma=6.
  2. 2.αβ+βγ+γα=5\alpha\beta+\beta\gamma+\gamma\alpha=5.
  3. 3.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).

Answer: 3610=2636-10=26.

Common mistakes

  • Don't fall into the trap of reading the sum of roots directly as the coefficient of xn1x^{n-1} without changing its sign.
  • Don't fall into the trap of using coefficient relations before dividing a non-monic polynomial by its leading coefficient.
  • Don't fall into the trap of replacing the sum of squares by the square of the sum and omitting the pair-product correction.

Exam tip

Write the required elementary symmetric sums from the coefficients before manipulating the target expression.

Tier 1 · Easy

ORIGINAL

1.

The roots of x35x2+2x+8=0x^3-5x^2+2x+8=0 are α,β,γ\alpha,\beta,\gamma. Without solving the equation, find α+β+γ\alpha+\beta+\gamma, αβ+βγ+γα\alpha\beta+\beta\gamma+\gamma\alpha and αβγ\alpha\beta\gamma.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

The non-zero roots of 2x43x35x2+7x4=02x^4-3x^3-5x^2+7x-4=0 are α,β,γ,δ\alpha,\beta,\gamma,\delta. Without solving the equation, find α+β+γ+δ\alpha+\beta+\gamma+\delta and 1α+1β+1γ+1δ\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The roots of x34x2+px6=0x^3-4x^2+px-6=0 are α,β,γ\alpha,\beta,\gamma, and α2+β2+γ2=10\alpha^2+\beta^2+\gamma^2=10. Determine pp and then find α3+β3+γ3\alpha^3+\beta^3+\gamma^3 without solving the cubic.

(5)

(Total for Question 1 is 5 marks)

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CP-4.2

Form a polynomial equation whose roots are a linear transformation of the roots of a given polynomial equation (of at least cubic degree).

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • If f(t)=0f(t)=0 has roots αi\alpha_i and the required new roots are x=aαi+bx=a\alpha_i+b, with a0a\ne0, rearrange the transformation to express the old root as t=(xb)/at=(x-b)/a, then substitute this into f(t)=0f(t)=0.
  • Multiplying by a suitable power of aa clears denominators without changing the roots, after which the polynomial is expanded and collected in powers of xx.
  • For a shift, reverse the shift in the substitution; for a scaling, divide the new variable by the scale factor.
  • The final equation may be multiplied by any non-zero constant, but its degree and number of transformed roots must remain unchanged.
Worked example

The roots of t3+2t4=0t^3+2t-4=0 are α,β,γ\alpha,\beta,\gamma. Form an equation with roots 2α+32\alpha+3, 2β+32\beta+3, 2γ+32\gamma+3.

  1. 1.Let x=2t+3x=2t+3, so t=(x3)/2t=(x-3)/2.
  2. 2.Substitute: ((x3)/2)3+2((x3)/2)4=0((x-3)/2)^3+2((x-3)/2)-4=0.
  3. 3.Multiply by 88: (x3)3+8(x3)32=0(x-3)^3+8(x-3)-32=0.
  4. 4.Expand and collect terms.

Answer: x39x2+35x83=0x^3-9x^2+35x-83=0.

Common mistakes

  • Don't fall into the trap of substituting t=ax+bt=ax+b instead of rearranging the new-root transformation for the old root.
  • Don't fall into the trap of clearing a cubic denominator by aa rather than by a3a^3.
  • Don't fall into the trap of changing the polynomial degree while expanding the transformed equation.

Exam tip

Define the new root, rearrange for the old variable, and display that substitution before expanding.

Tier 1 · Easy

ORIGINAL

1.

The roots of t33t+1=0t^3-3t+1=0 are α,β,γ\alpha,\beta,\gamma. Form a polynomial equation whose roots are α+2,β+2,γ+2\alpha+2,\beta+2,\gamma+2.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

The roots of 2t3t2+4t3=02t^3-t^2+4t-3=0 are α,β,γ\alpha,\beta,\gamma. Form a polynomial equation whose roots are 3α13\alpha-1, 3β13\beta-1 and 3γ13\gamma-1.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The roots of t42t3+t2+3t1=0t^4-2t^3+t^2+3t-1=0 are α,β,γ,δ\alpha,\beta,\gamma,\delta. Form a polynomial equation whose roots are 23α2-3\alpha, 23β2-3\beta, 23γ2-3\gamma and 23δ2-3\delta.

(6)

(Total for Question 1 is 6 marks)

CP-4.3

Understand and use formulae for the sums of integers, squares and cubes and use these to sum other series.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The standard results are r=1nr=n(n+1)2\sum_{r=1}^n r=\dfrac{n(n+1)}2, r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^n r^2=\dfrac{n(n+1)(2n+1)}6 and r=1nr3=[n(n+1)2]2\sum_{r=1}^n r^3=\left[\dfrac{n(n+1)}2\right]^2.
  • To sum a polynomial expression in rr, expand it, split the summation term by term and substitute the appropriate formulae.
  • A constant term cc contributes cncn when there are nn terms.
  • Different lower or upper limits may require subtracting an initial partial sum.
  • Factor before expanding the final expression where possible, and verify a closed form with a small value such as n=1n=1 to expose limit or constant-term errors.
Worked example

Find a closed form for r=1n(2r2+3r1)\sum_{r=1}^n(2r^2+3r-1).

  1. 1.Split the sum as 2r2+3r12\sum r^2+3\sum r-\sum1.
  2. 2.Substitute the standard results to obtain n(n+1)(2n+1)3+3n(n+1)2n\dfrac{n(n+1)(2n+1)}3+\dfrac{3n(n+1)}2-n.
  3. 3.Use a common denominator and simplify.

Answer: n(4n2+15n+5)6\dfrac{n(4n^2+15n+5)}6.

Common mistakes

  • Don't fall into the trap of treating r=1n1\sum_{r=1}^n1 as 11 instead of nn.
  • Don't fall into the trap of using the square-sum formula for a cubic term.
  • Don't fall into the trap of changing the summation limits when splitting one sum into several sums.

Exam tip

Show the split into standard sums before substitution; this makes the method visible even if later algebra slips.

Tier 1 · Easy

ORIGINAL

1.

Evaluate r=120r(r+1)\sum_{r=1}^{20}r(r+1).

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Show that r=1n(3r22r+4)=n(2n2+n+7)2\sum_{r=1}^{n}(3r^2-2r+4)=\frac{n(2n^2+n+7)}2.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Prove that r=1nr(r+1)(2r+1)=n(n+1)2(n+2)2\sum_{r=1}^{n}r(r+1)(2r+1)=\frac{n(n+1)^2(n+2)}2. Hence evaluate the sum when n=15n=15.

(6)

(Total for Question 1 is 6 marks)

CP-4.4

Understand and use the method of differences for summation of series including use of partial fractions.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The method of differences rewrites the general term as a difference such as f(r)f(r+1)f(r)-f(r+1) or f(r)f(r+k)f(r)-f(r+k). When consecutive terms are written out, most contributions cancel and only boundary terms remain.
  • Partial fractions often reveal this structure; the coefficients and index shift must be found exactly.
  • Write enough terms at the beginning and end of the finite sum to show the cancellation pattern, especially when the shift is greater than one.
  • The surviving first and last terms give the closed form.
  • For an infinite sum, first obtain the finite partial sum and then take its limit; cancellation is not a substitute for checking convergence.
Worked example

Find r=1n1(2r1)(2r+1)\sum_{r=1}^n\dfrac1{(2r-1)(2r+1)} by differences.

  1. 1.1(2r1)(2r+1)=12(12r112r+1)\dfrac1{(2r-1)(2r+1)}=\dfrac12\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right).
  2. 2.Writing the terms gives 12[(11/3)+(1/31/5)+]\dfrac12[(1-1/3)+(1/3-1/5)+\cdots].
  3. 3.All intermediate fractions cancel, leaving 12(11/(2n+1))\dfrac12(1-1/(2n+1)).

Answer: n2n+1\dfrac{n}{2n+1}.

Common mistakes

  • Don't fall into the trap of omitting the factor 1/21/2 from the partial-fraction decomposition.
  • Don't fall into the trap of cancelling a boundary term that has no matching term of opposite sign.
  • Don't fall into the trap of taking an infinite-series limit before deriving the finite partial sum.

Exam tip

For 'use the method of differences', display the first two and final two terms before stating what cancels.

Tier 1 · Easy

ORIGINAL

1.

Use the method of differences to find r=1n1(r+2)(r+3)\sum_{r=1}^{n}\frac1{(r+2)(r+3)}.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find a closed form for r=1n1r(r+2)\sum_{r=1}^{n}\frac1{r(r+2)}. Hence evaluate the sum for n=10n=10.

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

1.

Use partial fractions and the method of differences to prove that r=1n1r(r+1)(r+2)=n(n+3)4(n+1)(n+2)\sum_{r=1}^{n}\frac1{r(r+1)(r+2)}=\frac{n(n+3)}{4(n+1)(n+2)}. Hence find the corresponding infinite sum.

(6)

(Total for Question 1 is 6 marks)

CP-4.5

Find the Maclaurin series of a function including the general term.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A Maclaurin series is the Taylor series about zero: f(x)=r=0f(r)(0)r!xrf(x)=\sum_{r=0}^{\infty}\dfrac{f^{(r)}(0)}{r!}x^r, within the range of values of xx for which the expansion is valid.
  • Repeated differentiation supplies the derivative pattern and the values at zero.
  • A complete response should show enough initial terms to establish signs and missing powers, then state a general term with its index range.
  • Products and compound functions may instead be formed from known series, retaining every contribution up to the required degree; a complex exponential can efficiently generate some trigonometric products.
  • The factorial belongs to the denominator of every coefficient, and a finite list of terms does not replace a requested general term.
Worked example

Find the Maclaurin series for cosh(2x)\cosh(2x) and give its general term.

  1. 1.Successive derivatives alternate between multiples of cosh(2x)\cosh(2x) and sinh(2x)\sinh(2x).
  2. 2.Odd derivatives are zero at zero, while the 2r2rth derivative has value 22r2^{2r}.
  3. 3.Substitute these values into the Maclaurin formula.

Answer: cosh(2x)=r=022rx2r(2r)!=1+2x2+23x4+\cosh(2x)=\sum_{r=0}^{\infty}\dfrac{2^{2r}x^{2r}}{(2r)!}=1+2x^2+\dfrac23x^4+\cdots.

Common mistakes

  • Don't fall into the trap of giving several initial terms but no general term when one is explicitly requested.
  • Don't fall into the trap of using rr factorial for a series containing only powers x2rx^{2r} instead of (2r)!(2r)!.
  • Don't fall into the trap of including odd powers even though every odd derivative is zero at zero.

Exam tip

List derivative values at zero until the pattern is clear, then state the sigma term and its starting index.

Tier 1 · Easy

ORIGINAL

1.

Find the Maclaurin series of f(x)=11xf(x)=\frac1{1-x} and give its general term.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the Maclaurin series of e2te^{2t} up to and including the term in t4t^4, and state the general term.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Find the Maclaurin series of etcoste^t\cos t up to and including the term in t5t^5, and give a general term for the series.

(6)

(Total for Question 1 is 6 marks)

CP-4.6

Recognise and use the Maclaurin series for e^x, ln(1+x), sin x, cos x and (1+x)^n, and be aware of the range of values of x for which they are valid (proof not required).

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The standard series for exe^x, sinx\sin x and cosx\cos x converge for every real xx.
  • The logarithmic series is ln(1+x)=xx2/2+x3/3\ln(1+x)=x-x^2/2+x^3/3-\cdots, valid for 1<x1-1<x\le1.
  • The general binomial expansion begins (1+x)n=1+nx+n(n1)x2/2!+(1+x)^n=1+nx+n(n-1)x^2/2!+\cdots and, when it does not terminate, is valid for x<1|x|<1; a non-negative integer exponent produces a finite identity valid for all xx.
  • Compound expansions are obtained by substitution, multiplication or division of known series, keeping all terms up to the requested power.
  • Every substitution also changes the range of values of xx for which the expansion is valid, so the original condition must be rewritten.
Worked example

Write the first four non-zero terms of ln(12x)\ln(1-2x) and state the range of values of xx for which the expansion is valid.

  1. 1.Substitute u=2xu=-2x into ln(1+u)=uu2/2+u3/3u4/4+\ln(1+u)=u-u^2/2+u^3/3-u^4/4+\cdots.
  2. 2.This gives 2x2x28x3/34x4+-2x-2x^2-8x^3/3-4x^4+\cdots.
  3. 3.Transform 1<u1-1<u\le1 into 12x<1-1\le2x<1.

Answer: 2x2x283x34x4+-2x-2x^2-\dfrac83x^3-4x^4+\cdots, valid for 12x<12-\dfrac12\le x<\dfrac12.

Common mistakes

  • Don't fall into the trap of keeping the original interval after replacing xx by a multiple such as 2x-2x.
  • Don't fall into the trap of writing odd powers in the cosine series or even powers in the sine series.
  • Don't fall into the trap of treating a non-terminating binomial expansion as valid for all real xx.

Exam tip

State the substituted convergence condition alongside the expansion, not as an afterthought.

Tier 1 · Easy

ORIGINAL

1.

Write down the Maclaurin series for cost\cos t up to and including the term in t6t^6, and state the range of values of tt for which the expansion is valid.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Write the first four terms of the expansion of (12x)1/2(1-2x)^{-1/2}, and state the range of values of xx for which it is valid.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Use standard Maclaurin series to show that ln(1+x1x)=2(x+x33+x55+)\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right). State the range of values of xx for which the expansion is valid and use terms up to x5x^5 to approximate ln(3/2)\ln(3/2).

(6)

(Total for Question 1 is 6 marks)

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