The shape of the lesson
By the end of the lesson, learners can
- Algebraic long division: Polynomial division works like numerical long division.
- The remainder and factor theorems: Put \(x = -\dfrac{b}{a}\) into \(p(x) \equiv (ax + b)\,q(x) + r\): the bracket becomes zero, leaving \(r\).
- Finding unknown coefficients: Each piece of information (a factor, or a given remainder) gives one equation in the unknowns.
- Divide: leading term, multiply back, subtract, repeat; include zero terms for missing powers.
- Stop: when the remainder's degree is below the divisor's degree.
- Remainder theorem: remainder on dividing by \((ax + b)\) is \(p\left(-\dfrac{b}{a}\right)\).
- Factor theorem: \((ax + b)\) is a factor if and only if \(p\left(-\dfrac{b}{a}\right) = 0\).
- Unknowns: one equation per condition, then solve simultaneously.
One picture every idea returns to
\(p(x) = (\text{divisor}) \times (\text{quotient}) + \text{remainder}\), where the remainder has lower degree than the divisor. The remainder when \(p(x)\) is divided by \((ax + b)\) is \(p\left(-\dfrac{b}{a}\right)\), and \((ax + b)\) is a factor exactly when \(p\left(-\dfrac{b}{a}\right) = 0\).
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, Divide, then factorise.. Papers 2 and 3. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Algebraic long division | Polynomial division works like numerical long division. Model the worked example Dividing by a linear divisor on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The remainder and factor theorems | Put \(x = -\dfrac{b}{a}\) into \(p(x) \equiv (ax + b)\,q(x) + r\): the bracket becomes zero, leaving \(r\). Model the worked example Remainder for a non-unit divisor on the board, one line at a time. Your turn: Find the remainder when \(x^3 + 4x^2 - 7\) is divided by \((x + 3)\). | Pairs |
| 21 to 30 min | Teach and check: Finding unknown coefficients | Each piece of information (a factor, or a given remainder) gives one equation in the unknowns. Model the worked example Two conditions, two unknowns on the board, one line at a time. Your turn: \((3x + 2)\) is a factor of \(3x^3 + kx^2 - 7x - 2\). Find \(k\). | Think-Pair-Share |
| 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, 25 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 Finding unknown coefficients, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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 |
|---|---|
| Subtracting a negative term is where marks go: \((2x^2 + 5x) - (2x^2 - 4x) = 9x\), not \(x\). | Put the line you subtract in brackets and change every sign. |
| For the divisor \((x + 3)\) you substitute \(x = -3\), not \(x = 3\); for \((2x - 1)\) you substitute \(x = \dfrac{1}{2}\), not \(x = 1\) or \(x = 2\). | Return to The remainder and factor theorems: ask a pair to state the correct version and show it on the lab before moving on. |
| A remainder condition is \(p(c) = r\), not \(p(c) = 0\). | Only a factor gives zero. |
Polynomial division, factor and remainder theorems: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
Find the remainder when \(2x^3 - 3x^2 + x - 5\) is divided by \((x + 2)\).
Find the quotient and remainder when \(x^3 + 2x^2 - 5x + 1\) is divided by \((x - 3)\).
Find the quotient and remainder when \(2x^4 - x^3 + 3x^2 + 4\) is divided by \(x^2 - x + 2\).
(a) Show that \((3x + 1)\) is a factor of \(p(x) = 3x^3 - 2x^2 - 7x - 2\).
(b) Hence factorise \(p(x)\) completely and solve \(p(x) = 0\).
When \(x^3 + ax^2 - 4x + b\) is divided by \((x - 1)\) the remainder is \(-6\), and when it is divided by \((x + 2)\) the remainder is \(12\). Find \(a\) and \(b\).
The polynomial \(p(x) = 2x^3 + ax^2 + bx + 10\) has \((x + 2)\) as a factor. When \(p(x)\) is divided by \((x - 1)\) the remainder is \(-6\).
(a) Find the values of \(a\) and \(b\).
(b) Factorise \(p(x)\) completely.
(c) Hence solve \(2y^6 + ay^4 + by^2 + 10 = 0\), giving exact answers.
Answer key and marking notes
- \(p(-2) = 2(-8) - 3(4) + (-2) - 5 = -16 - 12 - 2 - 5 = -35\).
- Remainder \(-35\).
- \(x^2(x - 3) = x^3 - 3x^2\); subtract: \(5x^2 - 5x\).
- \(5x(x - 3) = 5x^2 - 15x\); subtract: \(10x + 1\).
- \(10(x - 3) = 10x - 30\); subtract: \(31\).
- Quotient \(x^2 + 5x + 10\), remainder \(31\). (Check: \(p(3) = 27 + 18 - 15 + 1 = 31\).)
- Write \(2x^4 - x^3 + 3x^2 + 0x + 4\).
- \(2x^2(x^2 - x + 2) = 2x^4 - 2x^3 + 4x^2\); subtract: \(x^3 - x^2 + 0x + 4\).
- \(x(x^2 - x + 2) = x^3 - x^2 + 2x\); subtract: \(-2x + 4\).
- Quotient \(2x^2 + x\), remainder \(-2x + 4\).
- (a) \(p\left(-\dfrac{1}{3}\right) = 3\left(-\dfrac{1}{27}\right) - 2\left(\dfrac{1}{9}\right) + \dfrac{7}{3} - 2 = -\dfrac{1}{9} - \dfrac{2}{9} + \dfrac{21}{9} - \dfrac{18}{9} = 0\), so \((3x + 1)\) is a factor.
- (b) \(p(x) = (3x + 1)(x^2 - x - 2)\) (check: \(x^2\) terms \(-3x^2 + x^2 = -2x^2\); \(x\) terms \(-6x - x = -7x\)).
- \(x^2 - x - 2 = (x - 2)(x + 1)\), so \(p(x) = (3x + 1)(x - 2)(x + 1)\).
- \(x = -\dfrac{1}{3},\ 2\) or \(-1\).
- \(p(1) = 1 + a - 4 + b = -6\), so \(a + b = -3\).
- \(p(-2) = -8 + 4a + 8 + b = 12\), so \(4a + b = 12\).
- Subtract: \(3a = 15\), so \(a = 5\) and \(b = -8\).
- (a) \(p(-2) = -16 + 4a - 2b + 10 = 0\), so \(2a - b = 3\).
- \(p(1) = 2 + a + b + 10 = -6\), so \(a + b = -18\).
- Add: \(3a = -15\), so \(a = -5\) and \(b = -13\).
- (b) \(p(x) = 2x^3 - 5x^2 - 13x + 10 = (x + 2)(2x^2 - 9x + 5)\). Check: \(x^2\) terms \(-9x^2 + 4x^2 = -5x^2\); \(x\) terms \(5x - 18x = -13x\).
- \(2x^2 - 9x + 5\) has discriminant \(81 - 40 = 41\), so it does not factorise over the integers: \(p(x) = (x + 2)(2x^2 - 9x + 5)\).
- (c) Put \(x = y^2\): \(y^2 = -2\) (no real solutions) or \(y^2 = \dfrac{9 \pm \sqrt{41}}{4}\).
- Both values \(\dfrac{9 \pm \sqrt{41}}{4}\) are positive (\(\sqrt{41} < 9\)), so \(y = \pm\dfrac{\sqrt{9 + \sqrt{41}}}{2}\) or \(y = \pm\dfrac{\sqrt{9 - \sqrt{41}}}{2}\).
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.
Find the remainder when \(x^3 - 4x^2 + 2x + 7\) is divided by \((x - 3)\).
- A\(4\)
- B\(-62\)
- C\(7\)
- D\(0\)
Find the remainder when \(4x^3 - 2x^2 + 3x + 5\) is divided by \((2x - 1)\).
- A\(\dfrac{13}{2}\)
- B\(\dfrac{5}{2}\)
- C\(35\)
- D\(5\)
Given that \((x + 2)\) is a factor of \(x^3 + ax^2 - 5x + 6\), find the value of the constant \(a\).
- A\(a = -2\)
- B\(a = -1\)
- C\(a = 2\)
- D\(a = 3\)
Find the quotient and the remainder when \(2x^3 + 3x^2 - 5x + 4\) is divided by \((x + 2)\).
- AQuotient \(2x^2 - x - 3\), remainder \(10\)
- BQuotient \(2x^2 + 7x + 9\), remainder \(22\)
- CQuotient \(2x^2 - x - 3\), remainder \(-2\)
- DQuotient \(2x^2 - x - 3\), remainder \(4\)
The polynomial \(p(x) = x^3 + ax^2 + bx - 12\) has \((x - 2)\) as a factor, and leaves remainder \(-24\) when divided by \((x + 1)\). Find \(a\) and \(b\).
- A\(a = -3\), \(b = 8\)
- B\(a = 15\), \(b = -28\)
- C\(a = 21\), \(b = 32\)
- D\(a = -3\), \(b = -8\)
Given that \((x - 3)\) is a factor of \(p(x) = 2x^3 - x^2 - 13x - 6\), solve the equation \(p(x) = 0\).
- A\(x = 3\), \(x = -2\) or \(x = -\dfrac{1}{2}\)
- B\(x = -3\), \(x = 2\) or \(x = \dfrac{1}{2}\)
- C\(x = 3\), \(x = -2\) or \(x = -1\)
- D\(x = 3\) only
Find the quotient and the remainder when \(x^4 - 3x^3 + 2x^2 + x - 1\) is divided by \(x^2 - x + 2\).
- AQuotient \(x^2 - 2x - 2\), remainder \(3x + 3\)
- BQuotient \(x^2 - 4x + 4\), remainder \(5x - 9\)
- CQuotient \(x^2 - 2x + 2\), remainder \(3 - x\)
- DQuotient \(x^2 - 2x - 2\), remainder \(-3x - 3\)
When \(p(x) = x^3 + kx^2 - 7x + 4\) is divided by \((x - 1)\) and by \((x + 2)\), the remainders are equal. Find \(k\).
- A\(k = -4\)
- B\(k = 4\)
- C\(k = -\dfrac{8}{5}\)
- D\(k = 2\)
Quiz key
\(p(3) = 27 - 36 + 6 + 7 = 4\). The tempting \(-62\) is \(p(-3)\), using the number in the bracket with its sign. 7 is just the constant term, \(p(0)\), and 0 would mean \((x - 3)\) is a factor, which it is not.
\(2x - 1 = 0\) when \(x = \dfrac{1}{2}\). \(p\left(\dfrac{1}{2}\right) = 4\left(\dfrac{1}{8}\right) - 2\left(\dfrac{1}{4}\right) + \dfrac{3}{2} + 5 = \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{3}{2} + 5 = \dfrac{13}{2}\). The tempting \(\dfrac{5}{2}\) is \(p\left(-\dfrac{1}{2}\right)\), a sign slip. 35 is \(p(2)\), reading the 2 from the bracket instead of solving, and 5 is only the constant term.
\(p(-2) = -8 + 4a + 10 + 6 = 4a + 8 = 0\), so \(a = -2\). The tempting \(a = -1\) comes from \(p(2) = 0\): \(8 + 4a - 10 + 6 = 0\). \(a = 2\) is a sign slip in the final step, and \(a = 3\) treats \(-5(-2)\) as \(-10\): \(-8 + 4a - 10 + 6 = 0\).
\(2x^3 + 3x^2 - 5x + 4 = (x + 2)(2x^2 - x - 3) + 10\): the steps give \(2x^2\), then \(-x^2 \div x = -x\), then \(-3x \div x = -3\), leaving \(4 - (-6) = 10\). Check: \(p(-2) = -16 + 12 + 10 + 4 = 10\). The tempting \(2x^2 + 7x + 9\) with remainder 22 is division by \((x - 2)\). Remainder \(-2\) comes from \(4 - 6\) instead of \(4 - (-6)\), and 4 is just the constant term.
\(p(2) = 8 + 4a + 2b - 12 = 0\) gives \(2a + b = 2\). \(p(-1) = -1 + a - b - 12 = -24\) gives \(a - b = -11\). Adding: \(3a = -9\), so \(a = -3\) and \(b = 8\). Check: \(p(x) = x^3 - 3x^2 + 8x - 12\), \(p(2) = 8 - 12 + 16 - 12 = 0\). The tempting \(a = 15\), \(b = -28\) uses \(p(1) = -24\) for the remainder, and \(a = 21\), \(b = 32\) uses \(p(-2) = 0\) for the factor. \(b = -8\) is a sign slip when back-substituting.
\(p(x) = (x - 3)(2x^2 + 5x + 2) = (x - 3)(2x + 1)(x + 2)\). So \(x = 3\), \(x = -\dfrac{1}{2}\) or \(x = -2\). Check: \(p(-2) = -16 - 4 + 26 - 6 = 0\). The tempting \(-3, 2, \dfrac{1}{2}\) reads the roots off the factors with the wrong signs. \(x = -1\) comes from solving \(2x + 1 = 0\) as \(x = -1\), and "\(x = 3\) only" stops after the given factor.
\(x^2(x^2 - x + 2) = x^4 - x^3 + 2x^2\); subtracting leaves \(-2x^3 + 0x^2 + x - 1\). Then \(-2x(x^2 - x + 2) = -2x^3 + 2x^2 - 4x\); subtracting leaves \(-2x^2 + 5x - 1\). Then \(-2(x^2 - x + 2) = -2x^2 + 2x - 4\); subtracting leaves \(3x + 3\). Check at \(x = 1\): \(p(1) = 0\) and \((2)(-3) + 6 = 0\). The tempting \(x^2 - 4x + 4\), \(5x - 9\) divides by \(x^2 + x + 2\); \(x^2 - 2x + 2\), \(3 - x\) divides by \(x^2 - x - 2\); and \(-3x - 3\) subtracts the wrong way round at the last step.
\(p(1) = 1 + k - 7 + 4 = k - 2\) and \(p(-2) = -8 + 4k + 14 + 4 = 4k + 10\). Equal: \(k - 2 = 4k + 10\), so \(3k = -12\), \(k = -4\). Check: \(p(1) = -6\) and \(p(-2) = -6\). The tempting \(k = 4\) uses \(p(-1)\) and \(p(2)\) (signs of the brackets copied). \(-\dfrac{8}{5}\) sets the remainders' SUM to zero, and \(k = 2\) just makes \(p(1) = 0\).
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.