The shape of the lesson
By the end of the lesson, learners can
- Choose the form first: Everything depends on writing down the correct form before you calculate anything.
- Three linear factors: With three different linear factors there are three constants, and each one comes from substituting the value of \(x\) that makes its own factor zero.
- A repeated factor: For \((cx + d)^2\) you need two fractions, \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\).
- An irreducible quadratic factor: When the denominator contains \(cx^2 + d\) (with \(c, d > 0\), so it has no real roots), its fraction needs a linear numerator \(Bx + C\).
- Numerator of equal degree: add a constant: When the numerator has the SAME degree as the denominator, the fraction is improper: it contains a whole-number part.
- Factorise the denominator fully, and compare degrees: equal degrees need a constant term \(A\) as well.
- Form: one fraction per linear factor; \(\dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^2}\) for a repeated factor; \(\dfrac{Bx + C}{cx^2 + d}\) for a quadratic factor.
- Identity: multiply through by the whole denominator.
- Substitute the root of each linear factor to find its constant directly.
- Compare coefficients (usually \(x^2\) and constants) for the rest, then check one more coefficient.
One picture every idea returns to
Choose the form from the denominator, multiply through to get an identity in \(x\), then find the constants by substituting the root of each linear factor and comparing coefficients for whatever is left. A repeated factor \((cx + d)^2\) needs BOTH \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\); a quadratic factor \(cx^2 + d\) needs a numerator \(Bx + C\). If the numerator has the same degree as the denominator, include a constant term as well.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One fraction, split.. Paper 3 only. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Choose the form first | Everything depends on writing down the correct form before you calculate anything. Model the worked example Two distinct linear factors on the board, one line at a time. Your turn: Express \(\dfrac{3x + 11}{x^2 - 2x - 3}\) in partial fractions. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Three linear factors | With three different linear factors there are three constants, and each one comes from substituting the value of \(x\) that makes its own factor zero. Model the worked example Three distinct factors, one non-unit on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: A repeated factor | For \((cx + d)^2\) you need two fractions, \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\). Model the worked example Repeated linear factor on the board, one line at a time. Your turn: Express \(\dfrac{x^2 + 10x + 7}{(x - 1)(x + 2)^2}\) in partial fractions. | Pairs |
| 23 to 30 min | Teach and check: the remaining sections | Sections 4 to 5 together: An irreducible quadratic factor; Numerator of equal degree: add a constant. Keep each to one idea and one check. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 29 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Numerator of equal degree: add a constant, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Factorise the denominator before choosing the form. \(x^2 - 2x - 3\) is not a 'quadratic factor' needing \(Bx + C\): it is \((x - 3)(x + 1)\). | Only a quadratic that does not factorise, such as \(x^2 + 3\), gets the \(Bx + C\) numerator. |
| When substituting \(x = -\dfrac{1}{2}\), every bracket must be evaluated at \(-\dfrac{1}{2}\): \((x - 1) = -\dfrac{3}{2}\), not \(-1\). | A single slip here makes \(C\) wrong, so always check with a coefficient comparison. |
| Writing only \(\dfrac{A}{x + 1} + \dfrac{C}{(2x + 1)^2}\) loses the middle term. | With two unknowns the identity cannot hold for every \(x\), and you will find the 'answers' disagree when you check. |
| \(\dfrac{B}{x^2 + 3}\) with a constant numerator is the wrong form: it gives too few unknowns, and the \(x\) coefficients will not match. | Over an irreducible quadratic always write \(Bx + C\). |
| Leaving out the constant term is the classic error here. | Without it the identity \(2x^2 + x + 3 \equiv B(2x + 1) + C(x - 1)\) is impossible: the right side has no \(x^2\) term. Compare the degrees of the numerator and denominator before you write down the form. |
Partial fractions: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
Write down the form of the partial fractions for each of the following. Do not find the constants.
(a) \(\dfrac{x + 1}{(x - 2)(3x + 1)}\) (b) \(\dfrac{2x^2 + 1}{(x + 4)^2(x - 5)}\) (c) \(\dfrac{x}{(2x - 1)(x^2 + 5)}\)
Express \(\dfrac{x^2 + 4x - 2}{(x - 2)(x + 3)}\) in partial fractions.
Express \(\dfrac{9x^2 + 5x - 10}{(x - 1)(x + 3)(2x - 1)}\) in partial fractions.
Express \(\dfrac{4 + 6x - x^2}{(2x - 1)(x + 1)^2}\) in partial fractions.
Express \(\dfrac{5x^2 + 11x}{(3x - 1)(x^2 + 2)}\) in partial fractions.
Let \(f(x) = \dfrac{x^2 + 8x + 11}{(x + 1)^2(x + 3)}\).
(a) Express \(f(x)\) in partial fractions.
(b) Hence find \(f'(x)\), and show that \(f'(0) = -\dfrac{53}{9}\).
Answer key and marking notes
- (a) \(\dfrac{A}{x - 2} + \dfrac{B}{3x + 1}\)
- (b) \(\dfrac{A}{x - 5} + \dfrac{B}{x + 4} + \dfrac{C}{(x + 4)^2}\)
- (c) \(\dfrac{A}{2x - 1} + \dfrac{Bx + C}{x^2 + 5}\)
- Numerator and denominator both have degree 2, so the form is \(A + \dfrac{B}{x - 2} + \dfrac{C}{x + 3}\).
- \(x^2 + 4x - 2 \equiv A(x - 2)(x + 3) + B(x + 3) + C(x - 2)\). Compare \(x^2\): \(A = 1\).
- \(x = 2\): \(10 = 5B\), so \(B = 2\). \(x = -3\): \(-5 = -5C\), so \(C = 1\).
- \(1 + \dfrac{2}{x - 2} + \dfrac{1}{x + 3}\)
- \(9x^2 + 5x - 10 \equiv A(x + 3)(2x - 1) + B(x - 1)(2x - 1) + C(x - 1)(x + 3)\).
- \(x = 1\): \(4 = A(4)(1)\), so \(A = 1\).
- \(x = -3\): \(81 - 15 - 10 = 56 = B(-4)(-7) = 28B\), so \(B = 2\).
- \(x = \dfrac{1}{2}\): \(\dfrac{9}{4} + \dfrac{5}{2} - 10 = -\dfrac{21}{4} = C\left(-\dfrac{1}{2}\right)\left(\dfrac{7}{2}\right) = -\dfrac{7}{4}C\), so \(C = 3\).
- \(\dfrac{1}{x - 1} + \dfrac{2}{x + 3} + \dfrac{3}{2x - 1}\) (check \(x^2\): \(2A + 2B + C = 2 + 4 + 3 = 9\)).
- \(4 + 6x - x^2 \equiv A(x + 1)^2 + B(2x - 1)(x + 1) + C(2x - 1)\).
- \(x = \dfrac{1}{2}\): \(4 + 3 - \dfrac{1}{4} = \dfrac{27}{4} = \dfrac{9}{4}A\), so \(A = 3\).
- \(x = -1\): \(4 - 6 - 1 = -3 = -3C\), so \(C = 1\).
- \(x^2\): \(-1 = A + 2B\), so \(B = -2\).
- \(\dfrac{3}{2x - 1} - \dfrac{2}{x + 1} + \dfrac{1}{(x + 1)^2}\)
- \(5x^2 + 11x \equiv A(x^2 + 2) + (Bx + C)(3x - 1)\).
- \(x = \dfrac{1}{3}\): \(\dfrac{5}{9} + \dfrac{11}{3} = \dfrac{38}{9} = A\left(\dfrac{19}{9}\right)\), so \(A = 2\).
- \(x^2\): \(5 = A + 3B\), so \(B = 1\).
- Constants: \(0 = 2A - C\), so \(C = 4\).
- \(\dfrac{2}{3x - 1} + \dfrac{x + 4}{x^2 + 2}\) (check \(x\): \(3C - B = 12 - 1 = 11\)).
- (a) \(x^2 + 8x + 11 \equiv A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2\), with \(f(x) = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x + 3}\).
- \(x = -1\): \(4 = 2B\), so \(B = 2\). \(x = -3\): \(-4 = 4C\), so \(C = -1\).
- \(x^2\): \(1 = A + C\), so \(A = 2\). So \(f(x) = \dfrac{2}{x + 1} + \dfrac{2}{(x + 1)^2} - \dfrac{1}{x + 3}\).
- (b) Write \(f(x) = 2(x + 1)^{-1} + 2(x + 1)^{-2} - (x + 3)^{-1}\) and differentiate each term with the chain rule.
- \(f'(x) = -\dfrac{2}{(x + 1)^2} - \dfrac{4}{(x + 1)^3} + \dfrac{1}{(x + 3)^2}\).
- \(f'(0) = -2 - 4 + \dfrac{1}{9} = -\dfrac{53}{9}\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
(Paper 3 only) Express \(\dfrac{5x + 1}{(x + 1)(x - 1)}\) in partial fractions.
- A\(\dfrac{2}{x + 1} + \dfrac{3}{x - 1}\)
- B\(\dfrac{3}{x + 1} + \dfrac{2}{x - 1}\)
- C\(-\dfrac{2}{x + 1} + \dfrac{3}{x - 1}\)
- D\(\dfrac{2}{x + 1} - \dfrac{3}{x - 1}\)
(Paper 3 only) Which is the correct form of partial fractions for \(\dfrac{x^2 + 1}{(2x - 1)(x^2 + 3)}\)?
- A\(\dfrac{A}{2x - 1} + \dfrac{Bx + C}{x^2 + 3}\)
- B\(\dfrac{A}{2x - 1} + \dfrac{B}{x^2 + 3}\)
- C\(\dfrac{A}{2x - 1} + \dfrac{B}{x + 3} + \dfrac{C}{(x + 3)^2}\)
- D\(A + \dfrac{B}{2x - 1} + \dfrac{Cx + D}{x^2 + 3}\)
(Paper 3 only) Express \(\dfrac{x^2 + 4x - 3}{(x + 1)(x - 2)}\) in partial fractions.
- A\(1 + \dfrac{2}{x + 1} + \dfrac{3}{x - 2}\)
- B\(\dfrac{2}{x + 1} + \dfrac{3}{x - 2}\)
- C\(x + \dfrac{2}{x + 1} + \dfrac{3}{x - 2}\)
- D\(1 - \dfrac{2}{x + 1} + \dfrac{3}{x - 2}\)
(Paper 3 only) Given that \(\dfrac{ax + b}{(x - 1)(x + 2)} \equiv \dfrac{3}{x - 1} - \dfrac{1}{x + 2}\), find the constants \(a\) and \(b\).
- A\(a = 2\), \(b = 7\)
- B\(a = 4\), \(b = 5\)
- C\(a = 2\), \(b = 5\)
- D\(a = 3\), \(b = -1\)
(Paper 3 only) Express \(\dfrac{7 - x}{(x + 2)(x - 1)^2}\) in partial fractions.
- A\(\dfrac{1}{x + 2} - \dfrac{1}{x - 1} + \dfrac{2}{(x - 1)^2}\)
- B\(\dfrac{1}{x + 2} + \dfrac{2}{(x - 1)^2}\)
- C\(\dfrac{1}{x + 2} + \dfrac{1}{x - 1} + \dfrac{2}{(x - 1)^2}\)
- D\(\dfrac{1}{x + 2} - \dfrac{1}{x - 1} + \dfrac{6}{(x - 1)^2}\)
(Paper 3 only) Express \(\dfrac{4x^2 + 3x - 12}{x(x - 2)(x + 3)}\) in partial fractions.
- A\(\dfrac{2}{x} + \dfrac{1}{x - 2} + \dfrac{1}{x + 3}\)
- B\(-\dfrac{2}{x} + \dfrac{1}{x - 2} + \dfrac{1}{x + 3}\)
- C\(\dfrac{2}{x} + \dfrac{1}{x - 2} - \dfrac{1}{x + 3}\)
- D\(\dfrac{2}{x} + \dfrac{5}{x - 2} + \dfrac{1}{x + 3}\)
(Paper 3 only) Express \(\dfrac{x^2 + 7}{(x + 1)(x^2 + 3)}\) in partial fractions.
- A\(\dfrac{2}{x + 1} + \dfrac{1 - x}{x^2 + 3}\)
- B\(\dfrac{2}{x + 1} + \dfrac{1}{x^2 + 3}\)
- C\(\dfrac{2}{x + 1} + \dfrac{x + 1}{x^2 + 3}\)
- D\(\dfrac{2}{x + 1} - \dfrac{x}{x^2 + 3}\)
(Paper 3 only) Express \(\dfrac{4x^2 + 3x + 2}{(2x - 1)(x + 1)}\) in partial fractions.
- A\(2 + \dfrac{3}{2x - 1} - \dfrac{1}{x + 1}\)
- B\(4 + \dfrac{3}{2x - 1} - \dfrac{1}{x + 1}\)
- C\(\dfrac{3}{2x - 1} - \dfrac{1}{x + 1}\)
- D\(2 + \dfrac{3}{2x - 1} + \dfrac{1}{x + 1}\)
Quiz key
\(x = 1\): \(6 = 2B\), so \(B = 3\). \(x = -1\): \(-4 = -2A\), so \(A = 2\). Answer \(\dfrac{2}{x + 1} + \dfrac{3}{x - 1}\). Check at \(x = 0\): \(2 - 3 = -1\) and \(\dfrac{1}{(1)(-1)} = -1\). The tempting \(\dfrac{3}{x + 1} + \dfrac{2}{x - 1}\) attaches each constant to the wrong factor (each constant goes with the factor that was NOT killed). The other two options are sign slips from dividing by \(-2\).
\(x^2 + 3\) is an irreducible quadratic, so its numerator must be linear, \(Bx + C\). The numerator \(x^2 + 1\) has degree 2, less than the degree 3 of the denominator, so no constant term is needed. The tempting \(\dfrac{B}{x^2 + 3}\) has too few constants: it cannot match every numerator. The repeated-factor form treats \(x^2 + 3\) as \((x + 3)^2\), which it is not, and the extra constant \(A\) is only needed when the degrees are equal.
\(x^2 + 4x - 3 \equiv A(x + 1)(x - 2) + B(x - 2) + C(x + 1)\). Comparing \(x^2\) coefficients: \(A = 1\). \(x = -1\): \(-6 = -3B\), \(B = 2\). \(x = 2\): \(9 = 3C\), \(C = 3\). The tempting \(\dfrac{2}{x + 1} + \dfrac{3}{x - 2}\) has the right values from the cover-up but forgets the constant; at \(x = 0\) it gives \(2 - \dfrac{3}{2} = \dfrac{1}{2}\), not \(\dfrac{-3}{-2} = \dfrac{3}{2}\). The constant is 1, not \(x\), because the degrees are equal, and \(-\dfrac{2}{x + 1}\) is a sign slip.
\(\dfrac{3(x + 2) - (x - 1)}{(x - 1)(x + 2)} = \dfrac{3x + 6 - x + 1}{(x - 1)(x + 2)} = \dfrac{2x + 7}{(x - 1)(x + 2)}\), so \(a = 2\), \(b = 7\). The tempting \(a = 4\), \(b = 5\) adds the second fraction instead of subtracting. \(b = 5\) comes from \(-(x - 1) = -x - 1\), and \(a = 3\), \(b = -1\) simply copies the two numerators.
\(7 - x \equiv A(x - 1)^2 + B(x + 2)(x - 1) + C(x + 2)\). \(x = 1\): \(6 = 3C\), \(C = 2\). \(x = -2\): \(9 = 9A\), \(A = 1\). \(x^2\) coefficients: \(0 = A + B\), so \(B = -1\). The tempting option without \(\dfrac{B}{x - 1}\) uses the wrong form; it cannot be right, since at \(x = 0\) it gives \(\dfrac{1}{2} + 2 = \dfrac{5}{2}\), but the original is \(\dfrac{7}{2}\). \(B = +1\) is a sign slip, and \(C = 6\) forgets to divide by \(x + 2 = 3\).
\(x = 0\): \(-12 = A(-2)(3)\), \(A = 2\). \(x = 2\): \(16 + 6 - 12 = 10 = B(2)(5)\), \(B = 1\). \(x = -3\): \(36 - 9 - 12 = 15 = C(-3)(-5)\), \(C = 1\). The tempting \(-\dfrac{2}{x}\) mishandles \(-12 \div (-6)\). \(-\dfrac{1}{x + 3}\) misses that \((-3)(-5)\) is positive, and \(\dfrac{5}{x - 2}\) divides 10 by 2 only, forgetting the factor \(x + 3 = 5\).
\(x^2 + 7 \equiv A(x^2 + 3) + (Bx + C)(x + 1)\). \(x = -1\): \(8 = 4A\), \(A = 2\). \(x^2\): \(1 = A + B\), so \(B = -1\). Constants: \(7 = 3A + C\), so \(C = 1\). Answer \(\dfrac{2}{x + 1} + \dfrac{1 - x}{x^2 + 3}\). The tempting \(\dfrac{1}{x^2 + 3}\) uses a constant numerator (wrong form) and only matches the constant terms. \(\dfrac{x + 1}{x^2 + 3}\) has \(B = +1\), a sign slip, and \(-\dfrac{x}{x^2 + 3}\) forgets \(C\).
\(4x^2 + 3x + 2 \equiv A(2x - 1)(x + 1) + B(x + 1) + C(2x - 1)\). \(x^2\): \(4 = 2A\), \(A = 2\). \(x = \dfrac{1}{2}\): \(1 + \dfrac{3}{2} + 2 = \dfrac{9}{2} = \dfrac{3}{2}B\), \(B = 3\). \(x = -1\): \(4 - 3 + 2 = 3 = -3C\), \(C = -1\). The tempting constant 4 copies the leading coefficient of the numerator without dividing by the 2 in \(2x^2\). Leaving out the constant gives the wrong function (at \(x = 0\): \(-3 - 1 = -4\), but the original is \(-2\)), and \(+\dfrac{1}{x + 1}\) is a sign slip.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.