Year 8 Factorising Worksheets with Answers

Worked examples

Year 8 factorising: worked examples

Four examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Finding the HCF of algebraic terms

   Find the highest common factor (HCF) of each pair or group of terms.

   (a) $12ab$ and $18a$

   (b) $20xy$, $15y$ and $10xyz$

   Show worked solution

   Key idea: The HCF of algebraic terms is the HCF of the numbers multiplied by every pronumeral that appears in all of the terms.

   1. (a) The factors of $12$ are 1, 2, 3, 4, 6, 12 and the factors of $18$ are 1, 2, 3, 6, 9, 18. The highest one they share is $6$.
   2. The pronumeral $a$ is in both terms. The pronumeral $b$ is only in $12ab$, so it is not part of the HCF.
      $\text{HCF}(12ab, 18a) = 6a$
   3. (b) The HCF of $20$, $15$ and $10$ is $5$.
   4. Only $y$ is in all three terms ($x$ is missing from $15y$, and $z$ is only in $10xyz$).
      $\text{HCF}(20xy, 15y, 10xyz) = 5y$

   Answer: (a) $6a$, (b) $5y$

2. Example 2 · Factorising by taking out the HCF

   Factorise fully.

   (a) $15x + 20$

   (b) $14m - 21mn$

   (c) $24n + 36$

   Show worked solution

   Key idea: Write the HCF outside a bracket. Inside the bracket, write what each term becomes when you divide it by the HCF. Then expand to check.

   1. (a) $\text{HCF}(15x, 20) = 5$. Divide each term: $15x \div 5 = 3x$ and $20 \div 5 = 4$.
      $15x + 20 = 5(3x + 4)$ Check: $5 \times 3x + 5 \times 4 = 15x + 20$ ✓
   2. (b) $\text{HCF}(14m, 21mn) = 7m$. Divide each term: $14m \div 7m = 2$ and $21mn \div 7m = 3n$.
      $14m - 21mn = 7m(2 - 3n)$ Check: $7m \times 2 - 7m \times 3n = 14m - 21mn$ ✓
   3. (c) $24n + 36$ can be written as $2(12n + 18)$ or $6(4n + 6)$, but neither is fully factorised because the bracket still has a common factor. The HCF of $24$ and $36$ is $12$.
      $24n + 36 = 12(2n + 3)$ Check: $12 \times 2n + 12 \times 3 = 24n + 36$ ✓, and $2n$ and $3$ have no common factor left.

   Answer: (a) $5(3x + 4)$, (b) $7m(2 - 3n)$, (c) $12(2n + 3)$

3. Example 3 · Taking out a negative common factor

   Factorise $-6p - 9q + 3$ by taking out a negative common factor.

   Show worked solution

   Key idea: When the first term is negative, take out the negative of the HCF. Dividing by a negative number changes the sign of every term inside the bracket.

   1. The HCF of $6$, $9$ and $3$ is $3$, so take out $-3$.
   2. Divide each term by $-3$:
      $\begin{gathered}-6p \div (-3) = 2p \\ -9q \div (-3) = 3q \\ 3 \div (-3) = -1\end{gathered}$
   3. Write the factorised form:
      $-6p - 9q + 3 = -3(2p + 3q - 1)$
   4. Check by expanding: $-3 \times 2p + (-3) \times 3q + (-3) \times (-1) = -6p - 9q + 3$ ✓

   Answer: $-3(2p + 3q - 1)$

4. Example 4 · Using factorising in a perimeter problem

   A rectangle has a perimeter of $(10w + 16)$ cm.

   (a) Factorise $10w + 16$ fully.

   (b) Use your answer to write an expression for the length plus the width.

   (c) The width is $(w + 3)$ cm. Find an expression for the length.

   Show worked solution

   Key idea: The perimeter of a rectangle is $2 \times (\text{length} + \text{width})$, so taking out the factor $2$ shows what length plus width must be.

   1. (a) $\text{HCF}(10w, 16) = 2$.
      $10w + 16 = 2(5w + 8)$
   2. (b) Comparing $2(5w + 8)$ with $2 \times (\text{length} + \text{width})$:
      $\text{length} + \text{width} = 5w + 8$
   3. (c) Subtract the width:
      $\text{length} = (5w + 8) - (w + 3) = 4w + 5$
   4. Check: $2\big((4w + 5) + (w + 3)\big) = 2(5w + 8) = 10w + 16$ ✓

   Answer: (a) $2(5w + 8)$, (b) $5w + 8$ cm, (c) $(4w + 5)$ cm

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