Algebraic Proof
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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.
Have a pen and paper handy, to work out your answers.
Question 12 marks
k is an integer.
A student is proving that (5k+2)(k+3)-(5k-1)(k+2) is always a multiple of 8
Which expression does (5k+2)(k+3)-(5k-1)(k+2) simplify to?
Hint
Expand each product separately, then put the second expansion in a bracket before you subtract.
Worked solution
- (5k+2)(k+3)=5k^2+15k+2k+6=5k^2+17k+6
- (5k-1)(k+2)=5k^2+10k-k-2=5k^2+9k-2
- Subtract, keeping the bracket: 5k^2+17k+6-(5k^2+9k-2)
- =5k^2+17k+6-5k^2-9k+2
- =8k+8
- =8(k+1), which is a multiple of 8
- Answer: 8k+8
Question 21 mark
n is an integer.
Which expression is always odd?
Hint
Try factorising the terms involving n and think about whether the product is odd or even when n is odd and when n is even.
Worked solution
- n^2+3n=n(n+3)
- If n is even, n(n+3) is even
- If n is odd, n+3 is even, so n(n+3) is even
- So n^2+3n is always even, and n^2+3n+5 is always odd
- The others are even when n is odd (e.g. n=1 gives 6, 8 and 8)
- Answer: n^2+3n+5
Question 3Challenge5 marks
n is an integer.
Expand and simplify (2n+7)^2-(2n-3)^2 Give your answer fully factorised.
3 marks
When n is odd, (2n+7)^2-(2n-3)^2 is always a multiple of m.
Work out the largest possible value of m.
2 marks
Hint
Bracket the second expansion before subtracting so every sign changes, then think about what n+1 is when n is odd.
Worked solution
Part (a)
- (2n+7)^2=4n^2+28n+49
- (2n-3)^2=4n^2-12n+9
- Subtract, keeping the bracket: 4n^2+28n+49-(4n^2-12n+9)
- =40n+40
- =40(n+1)
- (Or use the difference of two squares: (2n+7-2n+3)(2n+7+2n-3)=10(4n+4)=40(n+1))
Part (b)
- If n is odd, n+1 is even
- So n+1=2p for some integer p
- 40(n+1)=40\times 2p=80p, a multiple of 80
- n=1 gives 80 and n=3 gives 160, so no larger number always works
- m=80
More on this topic: Algebraic Proof worksheet with full solutions
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