Algebra: Factorisation By Common Factor & Grouping Practice 1
This section contains 8 carefully designed practice questions to help you strengthen your skills in factorising algebra using common factors and the grouping method. By working through these questions, you will gain more confidence in spotting patterns, breaking down expressions, and applying the rules step by step.
Related Lessons:
Practice 1 - Question 1
Factorise completely \(x^2y-3xy\)
Solution:
For this question, the common factor is \(xy\). Therefore, the answer is:
\(x^2y-3xy=xy\left(x-3\right)\)Practice 1 - Question 2
Factorise completely \(6ab-2a-5+15b\)
Solution:
This is a grouping factorisation since there are 4 terms. Always remember that the “sign” in the two brackets must always be the same. For this question, we need to swap the second bracket term.
\(6ab-2a-5+15b\)\(=2a\left(3b-1\right)-5\left(1-3b\right)\)
\(=2a\left(3b-1\right)+5\left(3b-1\right)\)
\(=\left(3b-1\right)\left(2a+5\right)\)
Practice 1 - Question 3
Factorise completely \(6ab-9+12b\)
Solution:
The common factor is only the numerical term, which is the value 3. Therefore, the factorisation form is:
\(6ab-9+12b=3\left(2ab-3+4b\right)\)Practice 1 - Question 4
Factorise completely \(2p^2+4pr-3pq-6qr\)
Solution:
Now, there are 4 terms in this expression, so we will perform grouping factorisation. Always remember that the “sign” in the two brackets must always be the same.
\(2p^2+4pr-3pq-6qr\)
\(=2p\left(p+2r\right)-3q\left(p+2r\right)\)
\(=\left(p+2r\right)\left(2p-3q\right)\)
Practice 1 - Question 5
Factorise completely \(-28cd+8d\)
Solution:
Now, some students may not be able to handle the factorisation if first term is a negative coefficient. In this case, move the positive number to the first term:
\[
\begin{align*}
-28cd+8d&=8d-28cd\\
&=4d\left(2-7c\right)\\
\end{align*}
\]
For those students who are able to handle the expression if the first algebraic term is negative, the answer is \(-4d\left(7c-2\right)\).
Practice 1 - Question 6
Factorise completely \(px-py+qy-qx\)
Solution:
Always remember that the “sign” in the two brackets must always be the same.
Note that we need to switch the term \(\left(y-x\right)\) to \(\left(x-y\right)\) since the bracket terms are not the same.
\(px-py+qy-qx\)\(=p\left(x-y\right)+q\left(y-x\right)\)
\(=p\left(x-y\right)-q\left(x-y\right)\)
\(=\left(x-y\right)\left(p-q\right)\)
Practice 1 - Question 7
Factorise completely \(12abc+4bcd\)
Solution:
This is a common factor factorisation, and the common factor is \(4bc\)
Practice 1 - Question 8
Factorise completely \(3xy-6x-y+2\)
Solution:
This is a grouping factorisation. 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.
\[
\begin{align*}
3xy-6x-y+2&=x\left(y-2\right)-1\left(y-2\right)\\
&=\left(y-2\right)\left(x-1\right)\\
\end{align*}
\]
Step By Step Full Solution
In Mathematics, arriving at the correct answer is important, but arriving at it using the correct method is even more essential. Below are the complete step-by-step solutions for the eight questions above. As students work through the full solutions, students may gain deeper understanding and uncover new insights.