Grade 12 Maths · Calculus · Paper 1

Remainder and Factor Theorem Grade 12

Real Remainder and Factor Theorem past paper questions and answers from 2021 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

past paper questions: 15 past paper questions
skill: 1 skill
videos: 1 videos

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Question 9.1 · May/June 2025 · Paper 1 · National (DBE) · 2 marks

Given:

f (x) = x^{3} - 6 x^{2} + 9 x - 4 = (x - 4) (x - k)^{2}

9.1 Show that

k = 1

.(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

Remainder and Factor Theorem skills

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Factorise a cubic polynomialFactorise a cubic by finding a value that makes it zero, then dividing to find the quadratic factor (for example x = 1 gives the factor x − 1).4 past paper questions · 4 videos

More Remainder and Factor Theorem Grade 12 questions and answers

Try each one on paper before you open the memo.

4. Question 1.1.1 · March 2023 · Provincial · 2 marks
1.1 Solve for $x$ :
1.1.1 $x (2 x - 3) = 0$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Solve for factors $x = 0$ or $2 x - 3 = 0$
  1 mark · Solve a quadratic by factorising
2. ✓
  Final answers $x = 0$ or $x = \frac{3}{2}$
  1 mark · Solve a quadratic by factorising
5. Question 1.1.1 · March 2022 · Provincial · 2 marks
1.1 Solve for $x$ :
1.1.1 $x^{2} - 7 x + 10 = 0$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Factorise the quadratic equation $(x - 5) (x - 2) = 0$
  1 mark · Factorise a trinomial
2. ✓
  Solve for x $x = 5$ or $x = 2$
  1 mark · Solve a quadratic by factorising
6. Question 1.1.1 · March 2021 · Provincial · 2 marks
1.1 Solve for $x$ :
1.1.1 $(x - 5) (x + 1) = 0$ (2)
Show the memo (1 tick)Memo
1. ✓
  Solve for x $x = 5$ or $x = - 1$
  2 marks · Solve a quadratic by factorising
7. Question WE13 · Siyavula Grade 12 Mathematics, 5.5 Solving cubic equations, Worked example 13 · 4 marks
WE13 Solve: $6 x^{3} - 5 x^{2} - 17 x + 6 = 0$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Find a factorLet $f (x) = 6 x^{3} - 5 x^{2} - 17 x + 6$
  $f (1) = 6 - 5 - 17 + 6 = - 10$ , so $(x - 1)$ is not a factor.
  $f (2) = 6 (2)^{3} - 5 (2)^{2} - 17 (2) + 6 = 48 - 20 - 34 + 6 = 0$ , so $(x - 2)$ is a factor.
  1 mark · Factorise a cubic polynomial
2. ✓
  Quadratic factor $6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) (6 x^{2} + 7 x - 3)$
  1 mark · Factorise a cubic polynomial
3. ✓
  Factorise fully $6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) (2 x + 3) (3 x - 1)$
  1 mark · Factorise trinomials with a leading coefficient
4. ✓
  Answer $(x - 2) (2 x + 3) (3 x - 1) = 0$
  $x = 2$ or $x = \frac{1}{3}$ or $x = - \frac{3}{2}$
  1 mark · Solve a quadratic by factorising

Remainder and Factor Theorem Grade 12 videos

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Factorise Cubic Equation Grade 12: IntroductionKevinmathscience

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