Grade 10 Maths · Algebra

Take out a common factor

Factorise by taking out the biggest common factor and putting the rest in a bracket (for example 6x² + 9x = 3x(2x + 3)).

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Gr 9 Factorisation: Common FactorKevinmathscience

Gr 9 Factorisation: Common Factor BracketKevinmathscience

FACTORISATION using the Highest Common Factor - Mathematics Grade 9SHFT - Your Learning

Grade 10 Maths Factorising Highest Common Factor (HCF)Miss Martins Maths and Science

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Question 1.3 · May/June 2025 · Paper 1 · National (DBE) · 3 marks

1.3 Show that

(1 + x^{m} + x^{- n})^{2} - (1 - x^{m} - x^{- n})^{2}

is divisible by

2

for all real values of

m

and

n

.(3)

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

More past paper questions: take out a common factor

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3. Question 1.1.1 · May/June 2024 · Paper 1 · National (DBE) · 2 marks
1.1 Solve for x:
1.1.1 $3 x^{2} + 5 x = 0$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Factorize the quadratic expression $x (3 x + 5) = 0$
  1 mark · Take out a common factor
2. ✓
  State both final solutions $x = 0$ or $x = - \frac{5}{3}$
  1 mark · Solve a quadratic by factorising
4. Question 6.2.3 · May/June 2025 · Paper 2 · National (DBE) · 5 marks
6.2 Consider the expression: $sin (A - B) - sin (A + B)$
6.2.3 Hence, determine the solution for $sin 4 x - sin 10 x = sin 3 x$ for $x \in [0^{\circ}; 3 0^{\circ}]$ .(5)
Show the memo (3 ticks)Memo
1. ✓
  Substitute the simplified term and factorize $- 2 cos 7 x sin 3 x = sin 3 x$
  $2 cos 7 x sin 3 x + sin 3 x = 0$
  $sin 3 x (2 cos 7 x + 1) = 0$
  2 marks · Take out a common factor
2. ✓
  Set each factor to zero $sin 3 x = 0$ OR $2 cos 7 x + 1 = 0$
  $sin 3 x = 0$ OR $cos 7 x = - \frac{1}{2}$
  1 mark · Solve a quadratic by factorising
3. ✓
  Solve for x in the given intervalFor $sin 3 x = 0$ :
  $3 x = 0^{\circ} + k \cdot 18 0^{\circ}$
  $x = 0^{\circ}$
  
  For $cos 7 x = - \frac{1}{2}$ (reference angle is $6 0^{\circ}$ , cosine is negative in 2nd quadrant):
  $7 x = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$
  $x = \frac{12 0 ^{\circ}}{7} \approx 17, 1 4^{\circ}$
  
  Other general solutions will fall outside the $[0^{\circ}; 3 0^{\circ}]$ domain restriction.
  
  $x = 0^{\circ}$ or $x = 17, 1 4^{\circ}$
  2 marks · Find solutions in an interval
5. Question 1.1.2 · November 2024 · Paper 1 · Provincial · 2 marks
1.1 Factorise the following expressions completely. Leave your answers in simplest form:
1.1.2 $a (6 a + 2 b) - 3 (3 a + b)$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Take out common factor in first term $2 a (3 a + b) - 3 (3 a + b)$
  1 mark · Take out a common factor
2. ✓
  Factorise by grouping $(3 a + b) (2 a - 3)$
  1 mark · Take out a common factor
6. Question 1.1.3 · March 2022 · Provincial · 3 marks
1.1 Consider: $9; 19; 33; 51; \dots$
1.1.3 Prove that all the terms of the quadratic sequence are odd.(3)
Show the memo (2 ticks)Memo
1. ✓
  Complete the square or factor $T_{n} = 2 (n^{2} + 2 n) + 3$
  $= 2 n (n + 2) + 3$
  OR
  $T_{n} = 2 (n + 1)^{2} + 1$
  2 marks · Take out a common factor
2. ✓
  Conclusion $2 (n + 1)^{2}$ is always even, adding 1 makes it odd.
  1 mark · Use a quadratic general term

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