← All notesMaths · Algebra: modulus, polynomials and partial fractions⬇ Download PDF
A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.1 · ~11 min read

Algebra: modulus, polynomials and partial fractions

The modulus function and its equations and inequalities, polynomial division, the factor and remainder theorems, partial fractions for the three syllabus forms, and the binomial expansion for any rational power with its range of validity.

Modulus graphsModulus equations & inequalitiesPolynomial divisionFactor theoremRemainder theoremPartial fractionsBinomial expansion (rational n)Range of validity
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • ǀxǀ is the distance from 0. Solve ǀaǀ = ǀbǀ by squaring (a² = b²), and ǀx − aǀ < b as a − b < x < a + b.
  • Remainder theorem: dividing p(x) by (ax + b) leaves remainder p(−b/a). Factor theorem: (ax + b) is a factor exactly when p(−b/a) = 0.
  • Partial fractions split a proper rational function into simpler fractions you can integrate or expand: one term per linear factor, two for a repeated factor, (Bx + C) over an irreducible quadratic.
  • (1 + x)ⁿ for rational n is an infinite series, valid only for ǀxǀ < 1. Adapt it by taking out a factor so the bracket starts with 1.

By the end you can

  • Sketch y = ǀax + bǀ and solve modulus equations and inequalities.
  • Divide a polynomial by a linear or quadratic divisor.
  • Use the factor and remainder theorems, including factors (ax + b).
  • Split rational functions into partial fractions of the three syllabus forms.
  • Expand (1 + x)ⁿ for rational n, adapt it, and state the set of x for which it is valid.

1The modulus function

Graphs and equations

ǀxǀ — the size of x without its sign: ǀxǀ = x if x ≥ 0, and ǀxǀ = −x if x < 0.

xy1234512345670(4, 5)(⅔, 5/3)1.5y = ǀ2x − 3ǀy = x + 1The V reflects the part of y = 2x − 3 below the axis. The line meets each arm once.
y = ǀ2x − 3ǀ is the line y = 2x − 3 with the part below the x-axis reflected upwards.
  • Sketching y = ǀax + bǀ: a V shape with its vertex on the x-axis at x = −b/a, y-intercept ǀbǀ. Paper 3 only needs linear expressions inside the modulus.
  • ǀaǀ = ǀbǀ ⇔ a² = b². Squaring removes both moduli without needing cases.
  • ǀaǀ = b (b an expression): solve a = b and a = −b, then check each answer, since b must be non-negative.
Worked example

Solve ǀ3x − 2ǀ = ǀ2x + 7ǀ.

  1. square both sides: 9x² − 12x + 4 = 4x² + 28x + 49 M1
  2. 5x² − 40x − 45 = 0 ⇒ x² − 8x − 9 = 0 ⇒ (x − 9)(x + 1) = 0 A1
  3. x = 9 or x = −1 A1
  4. check: ǀ25ǀ = ǀ25ǀ and ǀ−5ǀ = ǀ5ǀ ✓
Worked example

Solve ǀ2x − 3ǀ = x + 1.

  1. 2x − 3 = x + 1 ⇒ x = 4; −(2x − 3) = x + 1 ⇒ x = ⅔ M1 A1
  2. both give x + 1 > 0, so both are valid: x = 4 or x = ⅔ A1

Inequalities

ǀx − aǀ < b ⇔ a − b < x < a + b LEARN IT
and ǀx − aǀ > b ⇔ x < a − b or x > a + b
  • Two moduli: ǀf(x)ǀ < ǀg(x)ǀ ⇔ f(x)² < g(x)². Solve the quadratic inequality.
  • One modulus and a line: find the critical values (as for the equation), then use a sketch to choose the region.
Worked example

Solve 2x + 5 < ǀx + 1ǀ.

  1. for x ≥ −1: 2x + 5 < x + 1 ⇒ x < −4, which contradicts x ≥ −1: no solutions M1
  2. for x < −1: 2x + 5 < −x − 1 ⇒ x < −2 M1
  3. x < −2 A1 (check: x = −3 gives −1 < 2 ✓)
Common mistakeSquaring when only one side has a modulus, without checking. Squaring can introduce extra solutions; test each answer in the original equation.

2Polynomials

Division

  • Long division works like number division. Line up powers of x, including any missing powers with zero coefficients.
  • Result: p(x) = (divisor) × (quotient) + remainder, with the remainder of lower degree than the divisor.
Worked example

Divide 2x³ + x² − 13x + 6 by (x − 2), and hence factorise fully.

  1. 2x³ ÷ x = 2x²; subtract 2x²(x − 2) = 2x³ − 4x² to leave 5x² − 13x M1
  2. 5x² ÷ x = 5x; subtract 5x(x − 2) to leave −3x + 6; −3 × (x − 2) leaves 0 A1
  3. quotient 2x² + 5x − 3, remainder 0 A1
  4. so (x − 2)(2x − 1)(x + 3) A1

Factor and remainder theorems

p(x) ÷ (ax + b) leaves remainder p(−b/a) LEARN IT
remainder theorem; the factor theorem is the case remainder = 0
  • Why: p(x) = (ax + b)q(x) + R. Put x = −b/a and the first term vanishes, leaving R.
Worked example

p(x) = x³ + ax² + bx − 6. (x − 1) is a factor, and dividing by (x + 2) leaves −12. Find a and b.

  1. p(1) = 0 ⇒ 1 + a + b − 6 = 0 ⇒ a + b = 5 M1
  2. p(−2) = −12 ⇒ −8 + 4a − 2b − 6 = −12 ⇒ 2a − b = 1 M1
  3. a = 2, b = 3 A1 A1
Worked example

Show that (2x + 1) is a factor of 2x³ + 3x² − 11x − 6, and solve the equation = 0.

  1. p(−½) = 2(−⅛) + 3(¼) − 11(−½) − 6 = −¼ + ¾ + 5½ − 6 = 0, so (2x + 1) is a factor M1 A1
  2. 2x³ + 3x² − 11x − 6 = (2x + 1)(x² + x − 6) = (2x + 1)(x + 3)(x − 2) M1
  3. x = −½, −3, 2 A1
Common mistakeSolving a cubic on a calculator. The 2024 Paper 3 report: calculators may solve quadratics, but cubics and quartics must be factorised with working shown.

3Partial fractions

The three forms

DenominatorForm to use
(ax + b)(cx + d)(ex + f)A/(ax + b) + B/(cx + d) + C/(ex + f)
(ax + b)(cx + d)²A/(ax + b) + B/(cx + d) + C/(cx + d)²
(ax + b)(cx² + d)A/(ax + b) + (Bx + C)/(cx² + d)
  • The numerator must have lower degree than the denominator (in this syllabus it always will).
  • Finding constants: multiply through by the denominator, then substitute convenient values of x (the roots of each factor), and compare coefficients for any constant left.
Worked example

Express (5x + 1)/[(x − 1)(x + 2)] in partial fractions.

  1. 5x + 1 ≡ A(x + 2) + B(x − 1) M1
  2. x = 1: 6 = 3A ⇒ A = 2; x = −2: −9 = −3B ⇒ B = 3 A1
  3. 2/(x − 1) + 3/(x + 2) A1
Worked example

Express 4/[(x + 1)(x − 1)²] in partial fractions.

  1. 4 ≡ A(x − 1)² + B(x + 1)(x − 1) + C(x + 1) M1
  2. x = 1: 4 = 2C ⇒ C = 2; x = −1: 4 = 4A ⇒ A = 1 A1
  3. x² terms: 0 = A + B ⇒ B = −1 A1
  4. 1/(x + 1) − 1/(x − 1) + 2/(x − 1)²
Worked example

Express (3x² + x + 4)/[(x + 1)(x² + 1)] in partial fractions.

  1. 3x² + x + 4 ≡ A(x² + 1) + (Bx + C)(x + 1) M1
  2. x = −1: 6 = 2A ⇒ A = 3; x²: 3 = A + B ⇒ B = 0; constants: 4 = A + C ⇒ C = 1 A1 A1
  3. 3/(x + 1) + 1/(x² + 1)
Common mistakeUsing A/(x − 1) + B/(x − 1) for a repeated factor (it collapses to one fraction), or forgetting the Bx term over a quadratic. Both were common in 2024.

4The binomial expansion for rational n

The series

(1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + … , ǀxǀ < 1 FORMULA LIST (MF19)
for any rational n; an infinite series unless n is a positive integer
  • Adapting: make the bracket start with 1. (a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ, valid for ǀbx/aǀ < 1, i.e. ǀxǀ < ǀa/bǀ.
  • Brackets matter: in (1 − 4x)1/2, the "x" of the formula is (−4x), so x² becomes 16x² and x³ becomes −64x³.
Worked example

Expand √(1 − 4x) in ascending powers of x up to x³, and state the values of x for which it is valid.

  1. (1 + (−4x))1/2 = 1 + ½(−4x) + [(½)(−½)/2](−4x)² + [(½)(−½)(−3/2)/6](−4x)³ M1
  2. = 1 − 2x − 2x² − 4x³ A1 A1
  3. valid for ǀ4xǀ < 1, i.e. ǀxǀ < ¼ B1
Worked example

Expand (2 + x)⁻² up to x³.

  1. (2 + x)⁻² = 2⁻²(1 + x/2)⁻² = ¼[1 − 2(x/2) + 3(x/2)² − 4(x/2)³] M1
  2. = ¼ − x/4 + 3x²/16 − x³/8 A1
  3. valid for ǀxǀ < 2 B1
Worked example

Express f(x) = (5x + 1)/[(x − 1)(x + 2)] in partial fractions and hence expand f(x) up to x².

  1. f(x) = 2/(x − 1) + 3/(x + 2) B1
  2. 2/(x − 1) = −2(1 − x)⁻¹ = −2(1 + x + x² + …); 3/(x + 2) = (3/2)(1 + x/2)⁻¹ = 3/2 − 3x/4 + 3x²/8 M1 A1
  3. f(x) = −½ − 11x/4 − 13x²/8 + …, valid for ǀxǀ < 1 (the stricter condition) A1
Common mistakeExpanding (2 + x)⁻² as 2⁻²(1 + x)⁻², forgetting to divide x by 2 inside. Or using x instead of 2x, as in the 2024 Paper 3 report.
Linking questionPartial fractions are used to integrate in Further integration and to solve differential equations; the factor theorem gives complex roots in Complex numbers; and the positive-integer binomial theorem is in Series.

Test yourself — cover the answers first

Quick check

  1. Solve ǀx − 3ǀ < 5.
  2. What is the remainder when x³ − 2x + 4 is divided by (x + 1)?
  3. What form do you use for 1/[(x + 2)(x² + 3)]?
  4. For which x is the expansion of (1 + 3x)1/3 valid?
  5. Find the first two terms of (1 + x)⁻¹.

Exam-style

  1. Solve the inequality ǀx + 2ǀ > ǀ2x − 1ǀ.
  2. The polynomial 2x³ + ax² + bx + 3 has a factor (2x − 1) and leaves remainder 5 when divided by (x − 1). Find a and b.
  3. Expand (1 + 2x)/(1 − x)² in ascending powers of x up to x².
Show answers
  1. −2 < x < 8
  2. p(−1) = 5
  3. A/(x + 2) + (Bx + C)/(x² + 3)
  4. ǀxǀ < ⅓
  5. 1 − x
  6. (x + 2)² > (2x − 1)² ⇒ 3x² − 8x − 3 < 0 ⇒ (3x + 1)(x − 3) < 0 ⇒ −⅓ < x < 3
  7. p(½) = 0 ⇒ ¼ + a/4 + b/2 + 3 = 0 ⇒ a + 2b = −13; p(1) = 5 ⇒ a + b = 0 ⇒ b = −13, a = 13
  8. (1 + 2x)(1 + 2x + 3x² + …) = 1 + 4x + 7x² + …
Practise this chapter
A Level practice papers
Next chapterLogarithms and exponential functions →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.