Algebra: Factorisation By Common Factor & Grouping Practice 2
This practice set includes 8 focused questions that guide you through factorising using common factors and the grouping method. Each question helps you practise spotting patterns, simplifying expressions, and applying the steps clearly and confidently.
Related Lessons:
Practice 2 - Question 1
Factorise completely \(3ab-9a^2+6ac\)
Solution:
This is a common factor factorisation, and the common factor is \(3a\).
\(3ab-9a^2+6ac=3a\left(b-3a+2c\right)\)Practice 2 - Question 2
Factorise completely \(6ax-12by-8bx+9ay\).
Solution:
Always remember that the “sign” in the two brackets must always be the same. In this question, we will need to swap the term \((4b-3a)\) to \((3a-4b)\) since the bracket terms are not identical.
\(6ax-12by-8bx+9ay\)\(=6ax-8bx-12by+9ay\)
\(=2x\left(3a-4b\right)-3y\left(4b-3a\right)\)
\(=2x\left(3a-4b\right)+3y\left(3a-4b\right)\)
\(=\left(3a-4b\right)\left(2x+3y\right)\)
Practice 2 - Question 3
Factorise completely \(4\pi r^2+2\pi rh\)
Solution:
This is a common factor factorisation, and the common factor is \(2\pi r\)
\(4\pi r^2+2\pi rh=2\pi r\left(2r+h\right)\)Practice 2 - Question 4
Factorise completely \(6ac-21bc+35bd-10ad\)
Solution:
This is a grouping factorisation. Always remember that the “sign” in the two brackets must always be the same. And for this question, we need to swap \((7b-2a)\) to \((2a-7b)\):
\(6ac-21bc+35bd-10ad\)\(=3c\left(2a-7b\right)+5\left(7b-2a\right)\)
\(=3c\left(2a-7b\right)-5\left(2a-7b\right)\)
\(=\left(2a-7b\right)\left(3c-5d\right)\)
Practice 2 - Question 5
Factorise completely \(\left(x-2\right)^2-2\left(x-2\right)\)
Solution:
If you notice, the term \((x-2)\) is repeated between the 2 terms, so it is a common term:
\(\left(x-2\right)^2-2\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(x-2\right)-2\right]\)
\(=\left(x-2\right)\left(x-4\right)\)
If student expanded the term, student would need to perform cross factorisation (taught in Sec 2) to factorise the expression.
Practice 2 - Question 6
Factorise completely \(4eh+12hk-e-3k\)
Solution:
Always remember that the “sign” in the two brackets must always be the same. For this question, one common mistake students made is they miss out the value -1 in the second term.
\(4eh+12hk-e-3k\)
\(=4h\left(e+3k\right)-1\left(e+3k\right)\)
\(=\left(e+3k\right)\left(4h-1\right)\)
Practice 2 - Question 7
Factorise completely \(6a^2+12ab-6a\)
Solution:
This is a common factor factorisation and the common factor is \(6a\):
\(6a^2+12ab-6a=6a\left(a+2b-1\right)\)Practice 2 - Question 8
Factorise completely \(a-2b-ac+2bc\)
Solution:
Always remember that the “sign” in the two brackets must always be the same.
For this question, one common mistake students made is they miss out the value 1 in the first term.
\(a-2b-ac+2bc\)
\(=1\left(a-2b\right)-c\left(a-2b\right)\)
\(=\left(a-2b\right)\left(1-c\right)\)
Step By Step Full Solution
Below is the step-by-step video solutions for the eight practice questions above. It clearly demonstrates the complete working for each question, helping students reinforce their understanding and strengthen their Algebra skills.