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Non-calculator 1 markWhich one of these expressions is equivalent to \(x^2 + 7x + 12\)?
Write down the letter of your answer.A \((x + 12)(x + 1)\) B \((x + 3)(x + 4)\) C \((x - 3)(x - 4)\) D \((x + 2)(x + 5)\) Show answer
Option BFind two numbers that multiply to give \(12\) and add to give \(7\): they are \(3\) and \(4\).
\(x^2 + 7x + 12 = (x + 3)(x + 4)\), which is option B.
Check the others by expanding.
Option A, \((x + 12)(x + 1)\), gives \(x^2 + 13x + 12\). The product is right but the numbers add to \(13\), not \(7\).
Option C, \((x - 3)(x - 4)\), gives \(x^2 - 7x + 12\). The signs of both numbers are the wrong way round, so the middle term has the wrong sign.
Option D, \((x + 2)(x + 5)\), gives \(x^2 + 7x + 10\). The numbers add to \(7\) but multiply to \(10\), not \(12\).
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Non-calculator 2 marks
Expand and simplify \((3x - 2)(4x - 5)\)
Show answer
\(12x^2 - 23x + 10\)Multiply each term in the first bracket by each term in the second.
\(3x \times 4x = 12x^2\)
\(3x \times (-5) = -15x\)
\((-2) \times 4x = -8x\)
\((-2) \times (-5) = 10\)
Collect the two \(x\) terms: \(-15x - 8x = -23x\)
\(12x^2 - 23x + 10\)
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Non-calculator 2 marks
Multiply out the brackets and simplify \((3t - 4)^2\)
Show answer
\(9t^2 - 24t + 16\)Squaring a bracket means multiplying it by itself.
\((3t - 4)(3t - 4)\)
\(3t \times 3t = 9t^2\)
There are two cross terms, each \(3t \times (-4) = -12t\), so together they give \(-24t\).
\((-4) \times (-4) = 16\)
\(9t^2 - 24t + 16\)
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Non-calculator 2 marks
Factorise \(w^2 - 4w - 32\)
Show answer
\((w + 4)(w - 8)\)Look for two numbers that multiply to give \(-32\) and add to give \(-4\).
\(4 \times (-8) = -32\) and \(4 - 8 = -4\)
Those two numbers go straight into the brackets.
\(w^2 - 4w - 32 = (w + 4)(w - 8)\)
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Non-calculator 2 marks
\((t - 6)(t + a) = t^2 - 9t + 18\)
Calculate the value of \(a\).
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\(-3\)The two numbers in the brackets multiply to give the constant term and add to give the \(t\) coefficient. Either one finds \(a\).
Using the constant term: \((-6) \times a = 18\), so \(a = -3\)
Checking against the \(t\) term: \(-6 - 3 = -9\), which matches.
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Non-calculator 4 marks
Show, by expanding and simplifying, that \((4x - 7)(x - 6) + 3x(x + 1) - 7(x^2 - 4x + 4)\) is equal to an integer.
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The \(x^2\) terms and the \(x\) terms both cancel, leaving the integer \(14\).Expand each of the three terms on its own.
The pair of brackets: \((4x - 7)(x - 6) = 4x^2 - 24x - 7x + 42 = 4x^2 - 31x + 42\)
The term with \(x\) outside its bracket: \(3x(x + 1) = 3x^2 + 3x\)
The term with the multiplier outside its bracket: \(7(x^2 - 4x + 4) = 7x^2 - 28x + 28\)
Put the expansions back with their signs. The minus sign in front of the term with the multiplier outside its bracket changes the sign of every term inside it, so that term goes in a bracket:
\((4x^2 - 31x + 42) + (3x^2 + 3x) - (7x^2 - 28x + 28)\)
Remove the brackets:
\(4x^2 - 31x + 42 + 3x^2 + 3x - 7x^2 + 28x - 28\)
Collect the \(x^2\) terms: \((4 + 3 - 7)x^2 = 0\)
Collect the \(x\) terms: \((-31 + 3 + 28)x = 0\)
Collect the numbers: \(42 - 28 = 14\)
The \(x^2\) terms and the \(x\) terms cancel, so the expression simplifies to the integer \(14\).
- M1expands the pair of brackets to \(4x^2 - 31x + 42\)
- M1expands the term with \(x\) outside its bracket to \(3x^2 + 3x\)
- M1expands the trinomial bracket to \(7x^2 - 28x + 28\)
- A1applies the subtraction to every term of the term with the multiplier outside its bracket, shows the \(x^2\) and \(x\) terms cancel and states the integer 14
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Non-calculator 3 marks
\((x + a)(x - 4a) \equiv x^2 + bx - 16\)
Find the two possible values of \(b\).
Show answer
\(b = -6\) or \(b = 6\)Expand the brackets:
\((x + a)(x - 4a) = x^2 + a x - 4a x - 4a^2 = x^2 - 3a x - 4a^2\)
Compare the numbers term by term: \(b = -3a\) and \(-4a^2 = -16\).
So \(a^2 = 4\), which means \(a = 2\) or \(a = -2\). Both roots must be used.
When \(a = 2\): \(b = -3 \times 2 = -6\)
When \(a = -2\): \(b = -3 \times (-2) = 6\)
- M1expands to get \(x^2 - 3a x - 4a^2\), so \(-4a^2 = -16\)
- M1\(a^2 = 4\), so \(a = 2\) or \(a = -2\)
- A1both values of b: -6 and 6 (the positive root alone earns no mark)
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Non-calculator 3 marks
\(s\) and \(t\) are positive integers.
\((m - s)(m + t)\) expands and simplifies to \(m^2 - km - 80\), where \(k\) is a positive integer.
Work out the smallest possible value of \(k\).
Show answer
\(2\)Expanding: \((m - s)(m + t) = m^2 - (s - t)m - st\), so \(st = 80\) and \(k = s - t\).
For \(k\) to be positive, \(s\) must be larger than \(t\). The pairs of factors of 80 with \(s > t\) are
\(80 \times 1, 40 \times 2, 20 \times 4, 16 \times 5, 10 \times 8\)
Working out \(s - t\) for each pair gives \(k = 79, 38, 16, 11, 2\).
The smallest is \(k = 2\), from \(s = 10\) and \(t = 8\).
- M1expands the brackets to get \(st = 80\) and \(k = s - t\)
- M1lists the factor pairs of 80 with \(s > t\) and works out \(s - t\) for each, giving \(k = 79, 38, 16, 11, 2\)
- A1smallest value \(k = 2\)
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Non-calculator 1 markFactorise \(x^2 - 121\).
Write down the letter of your answer.A \((x - 11)^2\) B \(x(x - 121)\) C \((x + 11)^2\) D \((x + 11)(x - 11)\) Show answer
Option D\(x^2 - 121\) is a difference of two squares, and \(121 = 11^2\).
\(x^2 - 11^2 = (x + 11)(x - 11)\), which is option D.
Check by expanding. The cross terms \(+11x\) and \(-11x\) cancel and \(11 \times (-11) = -121\), so it gives \(x^2 - 121\).
Option C, \((x + 11)^2\), has the cross terms \(+11x\) twice, so it expands to \(x^2 + 22x + 121\).
Option A, \((x - 11)^2\), expands to \(x^2 - 22x + 121\).
Option B, \(x(x - 121)\), expands to \(x^2 - 121x\).
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Non-calculator 2 marks
Expand and simplify \((5x + 1)(4x + 7)\)
Show answer
\(20x^2 + 39x + 7\)Multiply each term in the first bracket by each term in the second.
\(5x \times 4x = 20x^2\)
\(5x \times 7 = 35x\)
\(1 \times 4x = 4x\)
\(1 \times 7 = 7\)
Collect the two \(x\) terms: \(35x + 4x = 39x\)
\(20x^2 + 39x + 7\)