Algebraic notation, expanding and factorising – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(3\times4-2\times(-3)=12+6=\) 18 (1 for substitution, 1 for the answer) Substitute with brackets: \(-2\times(-3)=+6\). So \(3p-2q=18\).
   2. \(3x^2+7x\) (1 for each correct term) Collect the \(x^2\) terms (\(5x^2-2x^2=3x^2\)) and the \(x\) terms (\(3x+4x=7x\)). The answer is \(3x^2+7x\) (2 marks).

   Check (a) another way: \(3p=12\) and \(2q=-6\), and subtracting a negative is the same as adding, so \(12-(-6)=18\). A common error is to write \(12-6=6\), which loses both marks. In (b), \(x^2\) and \(x\) are unlike terms, so \(3x^2+7x\) cannot be simplified any further.

2. Answer to Question 2 (4 marks)

   1. \(x^2-3x+7x-21=\) \(x^2+4x-21\) (1 for four terms, 1 for the answer) Collect the middle terms: \(-3x+7x=4x\).
   2. \((x-3)(x-5)\) (1) We need two numbers with product 15 and sum \(-8\): \(-3\) and \(-5\). Check: \(x^2-5x-3x+15\).
   3. \(3xy(2x-3y)\) (1) The HCF of 6 and 9 is 3 and both terms contain \(x\) and \(y\). Check: \(3xy\times2x=6x^2y\).

   In (a), test the expansion with \(x=1\): \((8)(-2)=-16\) and \(1+4-21=-16\), so the expansion is correct. In (c), taking out only 3 or only \(3x\) is not fully factorised and scores 0.

3. Answer to Question 3 (4 marks)

   1. \(2(3x+2)+2(x+4)=6x+4+2x+8=\) \(8x+12\) (1 for 2×sum, 1 for the simplified answer) The perimeter is two lengths plus two widths, so \(2(3x+2)+2(x+4)=8x+12\) metres, as required.
   2. \(8x+12=52\), \(x=5\) (1). Cost \(=52\times5=\) £260 (1) Solve \(8x=40\), so \(x=5\). The fence is 52 m long, so the total cost of the fence is £260.

   In (b), check the value of \(x\): the length is \(3\times5+2=17\) m and the width is \(5+4=9\) m, and \(2(17+9)=52\) m, which matches the perimeter given.

4. Answer to Question 4 (4 marks)

   1. \((2n+3)^2=4n^2+12n+9\) (1). \((2n-3)^2=4n^2-12n+9\) (1). Difference \(=24n\) (1). \(24n=8\times3n\), a multiple of 8 (1). Expanding both brackets and subtracting, the \(4n^2\) and \(9\) terms cancel. Since \(3n\) is a whole number, \((2n+3)^2-(2n-3)^2\) is a multiple of 8 for every whole number \(n\).

   A numerical check is useful but is not a proof: with \(n=1\), \(25-1=24=8\times3\), and with \(n=2\), \(49-1=48=8\times6\). Only the algebra, which works for every \(n\), earns the marks. Alternatively, use the difference of two squares: \((2n+3+2n-3)(2n+3-2n+3)=4n\times6=24n\).

5. Answer to Question 5 (4 marks)

   1. \(ac=-24\), numbers 3 and \(-8\): \(2x^2+3x-8x-12=x(2x+3)-4(2x+3)=\) \((2x+3)(x-4)\) (1 for split, 1 for the answer) We need two numbers with product \(2\times(-12)=-24\) and sum \(-5\). Check: \(2x^2-8x+3x-12=2x^2-5x-12\).
   2. \(x=-1.5\) or \(x=4\) (1) Set each bracket equal to zero: \(2x+3=0\) gives \(x=-1.5\) and \(x-4=0\) gives \(x=4\).
   3. \(\dfrac{(x-4)(x+4)}{(x+4)(x-3)}=\) \(\dfrac{x-4}{x-3}\) (1) The top is a difference of two squares and the bottom factorises with \(4\) and \(-3\); the common factor \((x+4)\) cancels.

Total : 20 marks