Further Mathematics · Pearson Edexcel 9FM0 · Topic 4

Further algebra and functions

Links between the roots and coefficients of polynomials, summing series with standard formulae and the method of differences, and Maclaurin series that approximate functions as polynomials.

  • 6 sub-topics
  • 3 in Year 12, 3 in Year 13
  • 99 practice questions
  • 6 animated lessons
Practice questions (99) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 4.1Roots and coefficients of polynomialsY12
  2. 4.2Transforming rootsY12
  3. 4.3Standard results for sumsY12
  4. 4.4Method of differencesY13
  5. 4.5Maclaurin seriesY13
  6. 4.6Standard Maclaurin seriesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 99 questions in this topic, free, no account needed.

Further algebra and functions

Easy

Roots of a quadratic · Year 12 · 3 marks

The roots of x2−5x+7=0x^2 - 5x + 7 = 0 are α\alpha and β\beta.
  1. (a)
    Write down α+β\alpha + \beta and αβ\alpha\beta.
    (1)
  2. (b)
    Find α2+β2\alpha^2 + \beta^2.
    (2)
Show worked solutionHide worked solution

Part (a)1 mark

  1. Sum =−ba= -\tfrac{b}{a}, product =ca= \tfrac{c}{a}.α+β\alpha + \beta=5, αβ{}= 5,\ \alpha\beta=7{}= 7

Part (b)2 marks

  1. Squaring the sum gives the squares plus 2αβ2\alpha\beta.α2+β2\alpha^2 + \beta^2=(α+β)2{}= (\alpha + \beta)^2−2αβ{}- 2\alpha\beta
  2. =25−14=11= 25 - 14 = 11

Total3

  • M
  • A
  • B
How did it go?

Report a problem with this question

Further algebra and functions

Medium

Roots of a cubic and their squares · Year 12 · 5 marks

The roots of 2x3−3x2+4x−5=02x^3 - 3x^2 + 4x - 5 = 0 are α\alpha, β\beta and γ\gamma.
  1. (a)
    Write down α+β+γ\alpha + \beta + \gamma, αβ\alpha\beta+βγ{}+ \beta\gamma+γα{}+ \gamma\alpha and αβγ\alpha\beta\gamma.
    (2)
  2. (b)
    Find α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 and explain what your answer tells you about the roots.
    (3)
Show worked solutionHide worked solution

Part (a)2 marks

  1. For ax3+bx2+cx+dax^3 + bx^2 + cx + d: ∑α=−ba\sum\alpha = -\tfrac{b}{a}, ∑αβ=ca\sum\alpha\beta = \tfrac{c}{a}, αβγ=−da\alpha\beta\gamma = -\tfrac{d}{a}.∑α\sum\alpha=32,{}= \tfrac{3}{2},∑αβ\quad \sum\alpha\beta=2,{}= 2,αβγ\quad \alpha\beta\gamma=52{}= \tfrac{5}{2}

Part (b)3 marks

  1. Squaring the sum of the roots gives the squares plus twice every pair.α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2=(∑α)2{}= (\sum\alpha)^2−2∑αβ{}- 2\sum\alpha\beta
  2. Substitute the values from part (a).=94−4=−74= \tfrac{9}{4} - 4 = -\tfrac{7}{4}
  3. Two of the roots must be a complex conjugate pair.The squares of real numbers cannot add to a negative, so not all the roots are real.

Total5

  • M
  • A
  • B
How did it go?

Report a problem with this question

Further algebra and functions

Hard

Roots of a quartic · Year 12 · 6 marks

The roots of x4+2x3−x+3=0x^4 + 2x^3 - x + 3 = 0 are α\alpha, β\beta, γ\gamma and δ\delta.
  1. (a)
    Find 1α\dfrac{1}{\alpha}+1β{}+ \dfrac{1}{\beta}+1γ{}+ \dfrac{1}{\gamma}+1δ{}+ \dfrac{1}{\delta}.
    (3)
  2. (b)
    Find α2\alpha^2+β2{}+ \beta^2+γ2{}+ \gamma^2+δ2{}+ \delta^2.
    (3)
Show worked solutionHide worked solution

Part (a)3 marks

  1. For ax4+bx3+cx2+dx+eax^4 + bx^3 + cx^2 + dx + e: ∑αβγ\sum\alpha\beta\gamma=−da{}= -\tfrac{d}{a} and αβγδ\alpha\beta\gamma\delta=ea{}= \tfrac{e}{a}.∑αβγ\sum\alpha\beta\gamma=1,{}= 1,αβγδ\quad \alpha\beta\gamma\delta=3{}= 3
  2. Put the four fractions over the common denominator αβγδ\alpha\beta\gamma\delta.∑1α\sum\dfrac{1}{\alpha}=∑αβγαβγδ{}= \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}
  3. Substitute the values.=13= \tfrac{1}{3}

Part (b)3 marks

  1. ∑α=−ba\sum\alpha = -\tfrac{b}{a} and ∑αβ=ca\sum\alpha\beta = \tfrac{c}{a}; there is no x2x^2 term, so c=0c = 0.∑α\sum\alpha=−2,{}= -2,∑αβ\quad \sum\alpha\beta=0{}= 0
  2. The same identity as for a cubic: it works for any number of roots.∑α2\sum\alpha^2=(∑α)2{}= (\sum\alpha)^2−2∑αβ{}- 2\sum\alpha\beta
  3. Substitute the values.=4−0=4= 4 - 0 = 4

Total6

  • M
  • A
  • B
How did it go?

Report a problem with this question