Introduction
This quadratic factorisation question is a very useful lower secondary algebra example because it tests both direct factorisation and the meaning of the word hence. Many students can do part (a), but get stuck in part (b) because they do not see how to reuse the earlier result. Once the pattern is recognised, this quadratic factorisation question becomes much easier and more systematic.
The Question / Scenario Explanation
Source: Factorisation(2) Sec 2 G3
Question (as shown):
(a) Factorise \(2x^2 – 11x – 21\).
(b) Hence, factorise \(2(2y+1)^2 – 22y – 32\).
Step-by-Step Solution / Explanation
Step 1: Factorise part (a)
We want to factorise:
\(2x^2 – 11x – 21\)
First multiply the coefficient of \(x^2\) and the constant term:
\(2 \times (-21) = -42\)
Now find two numbers that multiply to \(-42\) and add to \(-11\).
These numbers are \(-14\) and \(3\).
So split the middle term:
\(2x^2 – 14x + 3x – 21\)
Group the terms:
\(2x(x-7) + 3(x-7)\)
Factor out the common bracket:
\((2x+3)(x-7)\)
So part (a) gives:
\(2x^2 – 11x – 21 = (2x+3)(x-7)\)
Step 2: Use the word “hence” for part (b)
In this quadratic factorisation question, the word hence tells us to use the result from part (a).
Let:
\(x = 2y + 1\)
Now look at part (b):
\(2(2y+1)^2 – 22y – 32\)
Since \(x = 2y+1\), we also have:
\(2y = x – 1\)
So:
\(-22y = -11(2y) = -11(x-1)\)
Step 3: Rewrite part (b) in the same form as part (a)
Substitute into the expression:
\(2(2y+1)^2 – 22y – 32 = 2x^2 – 11(x-1) – 32\)
Simplify:
\(2x^2 – 11x + 11 – 32\)
\(= 2x^2 – 11x – 21\)
This is exactly the same expression as in part (a).
Step 4: Use the factorisation from part (a)
From part (a), we already know:
\(2x^2 – 11x – 21 = (2x+3)(x-7)\)
Now substitute back \(x = 2y+1\):
\((2(2y+1)+3)\big((2y+1)-7\big)\)
Simplify each bracket:
\((4y+2+3)(2y-6)\)
\((4y+5)(2y-6)\)
We can factorise further:
\((4y+5)(2y-6) = 2(4y+5)(y-3)\)
✅ Final Answers:
(a) \((2x+3)(x-7)\)
(b) \(2(4y+5)(y-3)\)
Step 5: Quick check for part (b)
Expand the final answer:
\(2(4y+5)(y-3)\)
\(= 2(4y^2 – 12y + 5y – 15)\)
\(= 2(4y^2 – 7y – 15)\)
\(= 8y^2 – 14y – 30\)
Now expand the original expression:
\(2(2y+1)^2 – 22y – 32\)
\(= 2(4y^2 + 4y + 1) – 22y – 32\)
\(= 8y^2 + 8y + 2 – 22y – 32\)
\(= 8y^2 – 14y – 30\) ✅
So the quadratic factorisation is correct.
Key Concepts Students Must Know
- In a quadratic factorisation question, always check whether the middle term can be split into two terms for grouping.
- The word hence means you should reuse the earlier result instead of starting from scratch.
- Substitution can help transform a new expression into one you have already factorised.
- Always simplify the final brackets fully and check whether there is any common factor left.
Exam Tips / Common Mistakes
Exam Tips
- For part (a), multiply the first coefficient and last term first to find the correct pair of numbers.
- In a quadratic factorisation question with the word “hence”, compare part (b) carefully with part (a).
- Write the substitution clearly, such as \(x = 2y+1\), so the working stays neat.
- Always simplify the final factorised form fully.
Common Mistakes
- Using the wrong pair of numbers for \(-42\) and \(-11\).
- Factorising part (a) correctly but not noticing that part (b) can be rewritten into the same form.
- Substituting \(x = 2y+1\) wrongly and making sign mistakes in the simplification.
- Stopping at \((4y+5)(2y-6)\) without spotting that it can also be written as \(2(4y+5)(y-3)\).
Parent Insight
This quadratic factorisation question is a good example of how secondary Maths rewards pattern recognition, not just calculation. Many students can factorise one quadratic, but they need practice seeing how one result can help solve another expression more efficiently. With guided practice, students become much stronger at spotting algebra links and working more confidently in exams.
Conclusion
To solve this quadratic factorisation question, we first factorised \(2x^2 – 11x – 21\) as \((2x+3)(x-7)\). Then we used the hint “hence” to rewrite part (b) in the same form and substituted back \(x = 2y+1\). This gave the final answers \((2x+3)(x-7)\) and \(2(4y+5)(y-3)\).
Visit our Secondary Maths homepage: Click Here…
Read more Secondary Maths worked examples here: Click Here…
Start here (limited slots weekly):
Free Trial: Click Here…
Enroll: Click Here…
Explore our main tuition brand approach here: Click Here… .