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Either paper 3 marks(a)
Expand \(7(4t + 3)\)
(b)Expand \(-7(t - 2)\)
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(a) \(28t + 21\)
(b) \(-7t + 14\)Multiply each term inside the bracket by 7.
\(7 \times 4t = 28t\) and \(7 \times 3 = 21\)
\(28t + 21\)
Both terms inside the bracket are multiplied by \(-7\), so both signs change.
\(-7 \times t = -7t\) and \(-7 \times (-2) = 14\)
\(-7t + 14\)
- (b) M1one term correct: \(-7t\) or \(14\)
- (b) A1both terms correct \(-7t + 14\)
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Either paper 2 marks(a)
Factorise fully \(10x - 2\)
(b)Factorise fully \(36x + 42y\)
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(a) \(2(5x - 1)\)
(b) \(6(6x + 7y)\)2 is the highest number that divides both 10 and 2.
\(10x \div 2 = 5x\) and \(2 \div 2 = 1\)
\(10x - 2 = 2(5x - 1)\)
6 is the highest number that divides both 36 and 42. The two terms have different letters, so no letter comes out of the bracket.
\(36x + 42y = 6(6x + 7y)\)
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Either paper 3 marks
Expand and simplify fully \(6(5y + 5) + 4(4y - 2)\)
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\(46y + 22\)Expand each bracket.
\(6(5y + 5) = 30y + 30\)
\(4(4y - 2) = 16y - 8\)
Now collect like terms.
\(y\) terms: \(30y + 16y = 46y\)
Numbers: \(30 - 8 = 22\)
\(46y + 22\)
- M1both brackets expanded \(30y + 30\) and \(16y - 8\)
- M1collecting like terms, \(y\) terms \(46y\) or numbers \(22\)
- A1simplified answer \(46y + 22\)
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Either paper 2 marks
Write \((x + 2)(x - 3)\) in the form \(x^2 + bx + c\), where \(b\) and \(c\) are integers.
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\(x^2 -x - 6\)Multiply each term in the first bracket by each term in the second.
\(x \times x = x^2\)
\(x \times (-3) = -3x\)
\(2 \times x = 2x\)
\(2 \times (-3) = -6\)
Collect the two \(x\) terms: \(-3x + 2x = -x\)
\(x^2 -x - 6\)
- M1expands the brackets with at least three terms correct, \(x^2\), \(-3x\), \(2x\), \(-6\)
- A1collects the middle terms \(x^2 -x - 6\)
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Either paper 2 marks
\(3(5k + c) = 15k + 12\)
Work out the value of \(c\).
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\(4\)Expanding multiplies both terms inside the bracket by 3, so the number on the right is \(3 \times c\).
\(3c = 12\)
\(c = 12 \div 3 = 4\)
Check: \(3 \times 4 = 12\), and \(3 \times 5k = 15k\)
- M1forms \(3c = 12\)
- A1\(c = 4\)
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Either paper 2 marks
\(12t + 16 = k(3t + 4)\)
Work out the value of \(k\).
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\(4\)Multiplying the bracket back out must give the left-hand side, so \(k\) times \(3t\) is \(12t\).
\(3k = 12\)
\(k = 12 \div 3 = 4\)
Check the constant as well: \(4 \times 4 = 16\)
- M1forms \(3k = 12\)
- A1\(k = 4\)
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Either paper 3 marks
Expand and simplify \(8(2m - 2) - 4(3m + 7)\)
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\(4m - 44\)Expand the first bracket.
\(8(2m - 2) = 16m - 16\)
The minus sign in front of the second bracket multiplies both terms inside it, so multiply by \(-4\) rather than by 4.
\(-4(3m + 7) = -12m - 28\)
Now collect like terms.
\(m\) terms: \(16m - 12m = 4m\)
Numbers: \(-16 - 28 = -44\)
\(4m - 44\)
- M1first bracket expanded \(16m - 16\)
- M1second bracket expanded with the sign changed \(-12m - 28\)
- A1collected like terms \(4m - 44\)
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Either paper 4 marks(a)
Factorise completely \(30m^2 - 42m\)
(b)Factorise completely \(48mn + 40m\)
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(a) \(6m(5m - 7)\)
(b) \(8m(6n + 5)\)Both terms contain \(m\), and 6 is the highest number that divides both 30 and 42. So the highest common factor is \(6m\).
\(30m^2 \div 6m = 5m\) and \(-42m \div 6m = -7\)
\(30m^2 - 42m = 6m(5m - 7)\)
Both terms contain \(m\), and 8 is the highest number that divides both 48 and 40. So the highest common factor is \(8m\).
\(48mn \div 8m = 6n\) and \(40m \div 8m = 5\)
\(48mn + 40m = 8m(6n + 5)\)
- (a) M1highest common factor \(6m\) taken out
- (a) A1fully factorised \(6m(5m - 7)\)
- (b) M1highest common factor \(8m\) taken out
- (b) A1fully factorised \(8m(6n + 5)\)
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Either paper 1 mark
Here is a student's attempt to expand and simplify:
\((x^2 + 3x - 12) - (x^2 -x - 10) = x^2 + 3x - 12 - x^2 -x - 10 = 2x - 22\)
Explain what has gone wrong in this working.
Show answer
When the bracket was subtracted, the sign of \(-x\) was not changed and the sign of \(- 10\) was not changed.Subtracting a whole bracket flips the sign of every term inside it, not only the first one.
\((x^2 -x - 10)\) becomes \(-x^2 +x + 10\) once it is subtracted.
In the working shown, the \(x^2\) term is flipped, the sign of \(-x\) was not changed, and the sign of \(- 10\) was not changed.
Done correctly, the expression simplifies to \(4x - 2\), not \(2x - 22\).
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Either paper 4 marks
\(m\) and \(n\) are integers.
(a)Prove that \((2m+1)^2 - (2n-1)^2 = 4(m+n)(m-n+1)\)
(b)Zainab says the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 12. Is Zainab correct? You must give reasons for your answer.
Show answer
(a) \((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n = 4(m+n)(m-n+1)\)
(b) No, Zainab is not correct: \(3\) and \(1\) are both odd, and \(3^2 - 1^2 = 9 - 1 = 8\), which is not a multiple of 12.Expand each square:
\((2m+1)^2 = 4m^2 + 4m + 1\)
\((2n-1)^2 = 4n^2 - 4n + 1\)
Subtract:
\((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n\)
Factorise fully:
\(4m^2 + 4m - 4n^2 + 4n = 4(m^2 + m - n^2 + n) = 4(m+n)(m-n+1)\)
Part (a) shows that \((2m+1)^2 - (2n-1)^2 = 4(m+n)(m-n+1)\) for any integers \(m\) and \(n\).
Taking \(m=1\) and \(n=1\) gives the odd numbers \(2m+1=3\) and \(2n-1=1\), and \(3^2 - 1^2 = 9 - 1 = 8\).
But \(8\) is not a multiple of 12, and Zainab is not correct.
- (a) M1expands both squares
- (a) M1simplifies to \(4m^2 + 4m - 4n^2 + 4n\)
- (a) A1factorises to \(4(m+n)(m-n+1)\)