Practise 36 exam-style questions on Expanding, factorising and simplifying for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
36 questions
1–5 marks each
Instant marking
No sign-up needed
20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Multiply every term inside each bracket by the number outside. The second bracket is multiplied by \(-3\), so \(-3 \times -4 = +12\). Writing \(-12\) here is the most common mistake.
Mark scheme
1 mark
M1 for expanding both brackets with at most one error, e.g. \(10y + 15 - 3y + 12\)
correct answer with no working scores full marks
1 mark
A1 for \(7y + 27\)
cao
Question 2
2 marks
Simplify \(\dfrac{12x^5y^3}{4x^2y^7}\)
Model answer
\(12 \div 4 = 3\) \(x^5 \div x^2 = x^3\) \(y^3 \div y^7 = y^{-4} = \dfrac{1}{y^4}\) So the expression simplifies to \(\dfrac{3x^3}{y^4}\)
Explanation
Divide the numbers and subtract the indices for each letter separately. For \(y\), \(3 - 7 = -4\), and a negative index means the power goes to the denominator. Writing \(y^4\) on top is a common slip.
Mark scheme
2 marks
B2 for \(\dfrac{3x^3}{y^4}\)
oe, e.g. \(3x^3y^{-4}\); B1 for two of the three parts correct, e.g. \(3x^3y^4\), \(\dfrac{3x^3}{y^{10}}\) or \(\dfrac{8x^3}{y^4}\)
The denominator is a difference of two squares, which tells you that \((2x + 1)\) or \((2x - 1)\) is likely to be a factor of the numerator too. Factorise both, then cancel the common factor \((2x - 1)\).
Mark scheme
1 mark
M1 for factorising the numerator, \((2x - 1)(x + 3)\)
correct answer with no working scores full marks
1 mark
M1 for factorising the denominator, \((2x + 1)(2x - 1)\)
1 mark
A1 for \(\dfrac{x + 3}{2x + 1}\)
cao
Question 4
4 marks
\(x\) and \(y\) are numbers such that \[\frac{2^x \times 8^y}{4^x} = 2^5\] and \[x + y = 7\]
Work out the value of \(x\) and the value of \(y\).
Model answer
\(\dfrac{2^x \times 2^{3y}}{2^{2x}} = 2^{x + 3y - 2x} = 2^{3y - x}\) So \(3y - x = 5\) Add \(x + y = 7\): \(4y = 12\), so \(y = 3\) \(x = 7 - 3 = 4\)
Explanation
Write every number as a power of 2, then use the index laws: add indices when multiplying and subtract when dividing. Comparing indices gives a second equation, which you solve with \(x + y = 7\). Check: \(\dfrac{16 \times 512}{256} = 32 = 2^5\).
Mark scheme
1 mark
M1 for writing \(8^y = 2^{3y}\) or \(4^x = 2^{2x}\)
correct answer with no working scores full marks
1 mark
M1 for \(x + 3y - 2x = 5\) or \(3y - x = 5\)
oe
1 mark
M1 for a correct method to solve their two equations simultaneously
e.g. adding to get \(4y = 12\)
1 mark
A1 for \(x = 4\) and \(y = 3\)
Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.
Ctrl + Enter to check
AI TutorHints first, answers after you try
Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.
