Factorising by grouping and special products questions with answers

Questions

1. Question 1Grouping: four terms

   Factorise $12x^{2} - 3x + 4xy - y$ completely.

   Show the worked solution

   1. Step 1: Group the terms in pairs and take out the HCF of each pair$12x^{2} - 3x + 4xy - y = 3x(4x - 1) + y(4x - 1)$
   2. Step 2: Take out the common bracket$= (4x - 1)(3x + y)$
2. Question 2Grouping: taking out a negative factor

   Factorise $mx + 7nx - my - 7ny$ completely.

   Show the worked solution

   1. Step 1: Group the terms in pairs and take out the HCF of each pair (a minus HCF changes the signs in its bracket)$mx + 7nx - my - 7ny = x(m + 7n) - y(m + 7n)$
   2. Step 2: Take out the common bracket$= (m + 7n)(x - y)$
3. Question 3Difference of two squares

   Factorise $81x^{2} - 64y^{2}$ completely.

   Show the worked solution

   1. Step 1: Write each term as a square$81x^{2} - 64y^{2} = (9x)^{2} - (8y)^{2}$
   2. Step 2: Use a² − b² = (a + b)(a − b)$= (9x + 8y)(9x - 8y)$
4. Question 4Common factor first, then a² − b²

   Factorise $24x^{2} - 150$ completely.

   Show the worked solution

   1. Step 1: Take out the HCF of the terms$24x^{2} - 150 = 6(4x^{2} - 25)$
   2. Step 2: Factorise the difference of two squares$= 6(2x + 5)(2x - 5)$
5. Question 5Common factor first, then a quadratic

   Factorise $36kx^{2} + 48kx + 16k$ completely.

   Show the worked solution

   1. Step 1: Take out the HCF of the terms first$36kx^{2} + 48kx + 16k = 4k(9x^{2} + 12x + 4)$
   2. Step 2: The bracket is a perfect square: (a + b)² = a² + 2ab + b²$= 4k(3x + 2)^{2}$
6. Question 6Quadratics in two letters

   Factorise $6m^{2} + 13mn - 5n^{2}$ completely.

   Show the worked solution

   1. Step 1: Find two numbers that multiply to 6 × (-5) and add to 13$-2 \times 15 = -30$ and $-2 + 15 = 13$
   2. Step 2: Split the middle term with them, group in pairs, and take out the common bracket$6m^{2} + 13mn - 5n^{2} = 6m^{2} - 2mn + 15mn - 5n^{2}$$= 2m(3m - n) + 5n(3m - n)$$= (2m + 5n)(3m - n)$
7. Question 7Hence: factorise by substitution

   Given that $3x^{2} - x - 10 = (3x + 5)(x - 2)$, factorise $3(y - 1)^{2} - (y - 1) - 10$ completely.

   Show the worked solution

   1. Step 1: It is the given expression with x replaced by y - 1: do the same in the brackets$3(y - 1)^{2} - (y - 1) - 10 = [3(y - 1) + 5][(y - 1) - 2]$
   2. Step 2: Expand inside each bracket and simplify$= (3y + 2)(y - 3)$
8. Question 8Grouping, then a² − b²

   Factorise $m^{2} - n^{2} + 5m + 5n$ completely.

   Show the worked solution

   1. Step 1: Factorise the first two terms (a difference of two squares) and take out the common factor of the last two$m^{2} - n^{2} + 5m + 5n = (m + n)(m - n) + 5(m + n)$
   2. Step 2: Take out the common bracket$= (m + n)(m - n + 5)$