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More questions on this topic: GCSE Algebra: Expanding Brackets (160) IGCSE Algebra: Expanding Brackets (136)
15 \(\quad (x - 3)(x + 4)(x + p) \equiv x^3 + qx^2 + rx + 24\)
Work out the values of \(p\), \(q\) and \(r\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(p = -2\) | B1 | |
| \(q = -1\) | B1ft | ft their \(p + 1\) or correct answer |
| \(r = -14\) | B1ft | ft their \(p - 12\) or correct answer |
| Fully correct expansion (could be seen in a grid) if no other marks scored \(x^3 + px^2 + x^2 + px - 12x - 12p\) oe | B1 |
8 \(k\) is a positive integer.
Prove that \(\quad 4k(k + 8) - (4k + 9)(k - 1) \quad\) is a multiple of 9 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4k^2 + 32k\) or \(4k^2 - 4k + 9k - 9\) or \(4k^2 + 5k - 9\) | M1 | may be seen in a full expansion eg \(\ldots - 4k^2 + 4k - 9k + 9\) may be seen in a grid |
| \(4k^2 + 32k - 4k^2 + 4k - 9k + 9\) or \(4k^2 + 32k - (4k^2 - 4k + 9k - 9)\) or \(4k^2 + 32k - 4k^2 - 5k + 9\) or \(4k^2 + 32k - (4k^2 + 5k - 9)\) | M1dep | correct 6-term expression and no errors or correct 5-term expression and no errors |
| M2 awarded and \(27k + 9\) and valid explanation involving 9 and \((3k + 1)\) | A1 | eg M2 awarded and \(27k + 9 = 9(3k + 1)\) or M2 awarded and \(\dfrac{27k + 9}{9} = 3k + 1\) |
| \(4k^2 - 4k + 9k - 9\) followed by an error in simplifying to 3 terms eg \(4k^2 - 4k + 9k - 9 = 4k^2 - 5k - 9 \quad (-5k\) is an error) Award 1st M1 but no further marks can be awarded | M1M0A0 |
| Do not award 2nd M1 if errors seen eg \(4k(k + 8) = 4k^2 + 32k\) \((4k + 9)(k - 1) = 4k^2 + 4k - 9k - 9 = 4k^2 + 5k - 9 \quad (+4k\) and \(-9k\) are errors) \(4k^2 + 32k - 4k^2 - 5k + 9\) (‘correct’ 5-term expression with errors) | M1 M0A0 |
| Missing brackets cannot be recovered | |
| For M marks accept eg \(- -9\) for \(+9\) | |
| Valid explanations involving 9 and \((3k + 1)\) include eg1 \(27k + 9\) when divided by 9 is \((3k + 1)\) eg2 \(27k + 9\) has factors \((3k + 1)\) and 9 | |
| Incomplete explanations include eg1 \(27k + 9\) which is divisible by 9 eg2 \(27k + 9\) and \(27k \div 9 = 3k\) and \(9 \div 9 = 1\) | |
| Condone eg \(x\) used for \(k\) |
21
\[\mathrm{f}(x) = \frac{2 - x}{3}\]\[\mathrm{g}(x) = 18x^2 + 15x\]\[\mathrm{f}^{-1}(x) + \mathrm{gf}(x) \quad \text{simplifies to} \quad ax^2 + bx + c\]Work out the values of \(a\), \(b\) and \(c\). [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2 - 3x\) or \(2 - 3y\) | B1 | oe accept \(2 - 3\mathrm{f}\) or \(2 - 3\mathrm{f}(x)\) |
| \(15\left(\dfrac{2 - x}{3}\right)\) or \(18\left(\dfrac{2 - x}{3}\right)^2\) | M1 | may be implied eg \(5(2 - x)\) or \(2(2 - x)^2\) |
| \(\dfrac{30 - 15x}{3}\) or \(10 - 5x\) | M1 | expansion of \(15\left(\dfrac{2 - x}{3}\right)\) implies 1st M1 |
| \(18\left(\dfrac{4 - 2x - 2x + x^2}{9}\right)\) or \(18\left(\dfrac{4 - 4x + x^2}{9}\right)\) or \(8 - 8x + 2x^2\) | M1 | oe squaring of bracket in \(18\left(\dfrac{2 - x}{3}\right)^2\) eg \(2(4 - 2x - 2x + x^2)\) or \(2(4 - 4x + x^2)\) implies 1st M1 |
| \(8 - 4x - 4x + 2x^2 + 10 - 5x\) or \(2x^2 - 13x + 18\) | M1dep | oe \(\mathrm{gf}(x)\) with elimination of brackets and fractions eg \(8 - 8x + 2x^2 + 10 - 5x\) dep on M3 \(2x^2 - 16x + 20\) is B1M4 |
| \(a = 2 \quad b = -16 \quad c = 20\) | A1ft | ft their \(2 - 3x\) if B0M4 their \(2 - 3x\) must be of the form \(px + q\) with \(p\) and \(q\) non-zero their \(2 - 3x\) cannot be \(\dfrac{2 - x}{3}\) |
| Missing brackets must be recovered | |
| Ignore irrelevant brackets eg \((8 - 4x - 4x + 2x^2) + (10 - 5x)\) | M4 |
| \(3x - 2\) \(8 - 4x - 4x + 2x^2 + 10 - 5x\) (\(3x - 2\) added to the above expression is \(2x^2 - 10x + 16\)) \(a = 2 \quad b = -10 \quad c = 16\) | B0 M4 A1ft |
13 Expand and simplify fully \(\quad (3x - 1)(4 - x)(2x + 5)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: expands \((3x - 1)(4 - x)\) first | ||
| Exactly 4 terms with at least 3 correct from \((+)12x \quad -3x^2 \quad -4 \quad (+)x\) | M1 | oe eg \(13x - 3x^2 - 4\) terms may be seen in a grid implied by \(13x - 3x^2 + k\) or \(13x + px^2 - 4\) where \(k\) and \(p\) are non-zero constants |
| Full expansion of their 3 or 4 term quadratic with \((2x + 5)\) with correct multiplication of their 3 or 4 terms by \(2x\) or by \((+)5\) | M1 | eg \(24x^2 - 6x^3 - 8x + 2x^2 + 60x - 15x^2 - 20 + 5x\) or \(26x^2 - 6x^3 - 8x + 65x - 15x^2 - 20\) terms may be seen in a grid |
| \(-6x^3 + 11x^2 + 57x - 20\) | A1 | terms in any order |
| Alternative method 2: expands \((4 - x)(2x + 5)\) first | ||
| Exactly 4 terms with at least 3 correct from \((+)8x \quad (+)20 \quad -2x^2 \quad -5x\) | M1 | oe eg \(3x + 20 - 2x^2\) terms may be seen in a grid implied by \(3x + 20 + kx^2\) or \(3x + p - 2x^2\) where \(k\) and \(p\) are non-zero constants |
| Full expansion of their 3 or 4 term quadratic with \((3x - 1)\) with correct multiplication of their 3 or 4 terms by \(3x\) or by \(-1\) (corrected from the printed mark scheme: printed as \((3x + 1)\)) | M1 | eg \(24x^2 + 60x - 6x^3 - 15x^2 - 8x - 20 + 2x^2 + 5x\) or \(9x^2 + 60x - 6x^3 - 3x - 20 + 2x^2\) terms may be seen in a grid |
| \(-6x^3 + 11x^2 + 57x - 20\) | A1 | terms in any order |
| Alternative method 3: expands \((3x - 1)(2x + 5)\) first | ||
| Exactly 4 terms with at least 3 correct from \((+)6x^2 \quad (+)15x \quad -2x \quad -5\) | M1 | oe eg \(6x^2 + 13x - 5\) terms may be seen in a grid implied by \(6x^2 + 13x + k\) or \(px^2 + 13x - 5\) where \(k\) and \(p\) are non-zero constants |
| Full expansion of their 3 or 4 term quadratic with \((4 - x)\) with correct multiplication of their 3 or 4 terms by 4 or by \(-x\) | M1 | eg \(24x^2 + 60x - 8x - 20 - 6x^3 - 15x^2 + 2x^2 + 5x\) or \(24x^2 + 52x - 20 - 6x^3 - 13x^2 + 5x\) terms may be seen in a grid |
| \(-6x^3 + 11x^2 + 57x - 20\) | A1 | terms in any order |
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone one transcription error for A1 after fully correct answer seen | |
| Alt 1 \(\ 12x + 3x^2 - 5 + x \quad\) only 2 terms correct \((12x + 3x^2 - 5 + x)(2x + 5)\) \(= 24x^2 + 6x^3 - 10x + 2x^2 + 17x + 15x^2 + 25 + 5x^2\) 8 terms (full expansion) with correct multiplication of their 4 terms by \(2x\) | M0 M1 |
| Alt 2 \(\ 3x + 20 + 2x^2 \quad\) implied 4 terms with 3 correct \((3x - 1)(3x + 20 + 2x^2)\) \(= 6x^2 + 60x + 5x^3 - 3x - 20 - 2x^2\) 6 terms (full expansion) with correct multiplication of their 3 terms by \(-1\) | M1 M1 |
| If the 1st M1 is scored with a 4-term expansion but this is then incorrectly simplified to 3 terms, the 2nd M1 can still be scored using their 3 terms | |
| Accept eg \(+ -3x^2\) as a correct term for M marks but not for A1 | |
| M2A1 response followed by further work eg1 \(-6x^3 + 11x^2 + 57x - 20\) in working with answer \(x(-6x^2 + 11x + 57) - 20\) eg2 \(-6x^3 + 11x^2 + 57x - 20\) in working with answer \(-(6x^3 - 11x^2 - 57x + 20)\) | M2A0 M2A0 |
| One single expansion is full marks or zero |
11 Expand and simplify fully \(\quad (2x^3 - 9)(3x^2 + 4) + x(x - 4)^2\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| 3 correct from \(6x^5 + 8x^3 - 27x^2 - 36\) | M1 | expansion of \((2x^3 - 9)(3x^2 + 4)\) may be seen in a grid |
| \(6x^5 + 8x^3 - 27x^2 - 36\) | M1dep | may be seen in a grid |
| \(x^3 - 4x^2 - 4x^2 + 16x\) or \(x^3 - 8x^2 + 16x\) | M1 | expansion of \(x(x - 4)^2\) may be seen in a grid |
| \(6x^5 + 9x^3 - 35x^2 + 16x - 36\) | A1 |
| A correct term includes the sign (in a grid allow eg \(8x^3\) for \(+8x^3\)) | |
| Terms may be in any order throughout | |
| Terms must be fully processed eg do not allow \(2x^3 \times 3x^2\) for \(6x^5\) | |
| Allow eg \(+ -27x^2\) for \(-27x^2\) for M marks | |
| \(6x^5 + 8x^3 - 27x^2 - 36\) scores the 1st M1 and 2nd M1 even if incorrectly simplified (or differentiated) eg \(6x^5 + 8x^3 - 27x^2 - 36 = 6x^5 - 19x^2 - 36\) | M1M1 |
| \(x^3 - 4x^2 - 4x^2 + 16x\) (or \(x^3 - 8x^2 + 16x\)) scores the 3rd M1 even if incorrectly simplified (or differentiated) eg \(x^3 - 4x^2 - 4x^2 + 16x = x^3 - 16x^2 + 16x\) | 3rd M1 |
| Correct answer then incorrect further work eg differentiation | M3A0 |
| \(6x^5 + 9x^3 - 35x^2 + 16x - 36 = x(6x^4 + 9x^2 - 35x + 16) - 36\) | M3A0 |
19 Expand and simplify fully \(\qquad (3x + 4)(2x - 3)(5x - 2)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Expands \((3x + 4)(2x - 3)\) first | ||
| \(6x^2 - 9x + 8x - 12\) or \(6x^2 - x - 12\) | M1 | oe 4 terms with at least 3 correct implied by \(6x^2 - x + k\) or \(px^2 - x - 12\) where \(k\) and \(p\) are non-zero constants may be seen in a grid |
| \(30x^3 - 45x^2 + 40x^2 - 60x - 12x^2\) \(+ 18x - 16x + 24\) or \(30x^3 - 5x^2 - 60x - 12x^2 + 2x + 24\) | M1 | oe full expansion with correct multiplication of their 3 or 4 terms by \(5x\) or \(-2\) may be seen in a grid |
| \(30x^3 - 17x^2 - 58x + 24\) | A1 | terms in any order |
| Alternative method 2 Expands \((2x - 3)(5x - 2)\) first | ||
| \(10x^2 - 4x - 15x + 6\) or \(10x^2 - 19x + 6\) | M1 | oe 4 terms with at least 3 correct implied by \(10x^2 - 19x + k\) or \(px^2 - 19x + 6\) where \(k\) and \(p\) are non-zero constants may be seen in a grid |
| \(30x^3 - 12x^2 - 45x^2 + 18x + 40x^2\) \(- 16x - 60x + 24\) or \(30x^3 - 57x^2 + 18x + 40x^2 - 76x\) \(+ 24\) | M1 | oe full expansion with correct multiplication of their 3 or 4 terms by \(3x\) or 4 may be seen in a grid |
| \(30x^3 - 17x^2 - 58x + 24\) | A1 | terms in any order |
| Alternative method 3 Expands \((3x + 4)(5x - 2)\) first | ||
| \(15x^2 - 6x + 20x - 8\) or \(15x^2 + 14x - 8\) | M1 | oe 4 terms with at least 3 correct implied by \(15x^2 + 14x + k\) or \(px^2 + 14x - 8\) where \(k\) and \(p\) are non-zero constants may be seen in a grid |
| \(30x^3 - 12x^2 + 40x^2 - 16x - 45x^2\) \(+ 18x - 60x + 24\) or \(30x^3 + 28x^2 - 16x - 45x^2 - 42x\) \(+ 24\) | M1 | oe full expansion with correct multiplication of their 3 or 4 terms by \(2x\) or \(-3\) may be seen in a grid |
| \(30x^3 - 17x^2 - 58x + 24\) | A1 | terms in any order |
| For terms seen in a grid accept \(8x\) for \(+8x\) etc | |
| 2nd M1 A full expansion will be 8 terms if 4 terms are used in first expansion A full expansion will be 6 terms if 3 terms are used in first expansion | |
| Alt 1 \(6x^2 + 9x - 8x - 12 \qquad\) only 2 terms correct \((6x^2 + 9x - 8x - 12)(5x - 2)\) \(= 30x^3 + 45x^2 - 40x^2 - 60x - 12x^2 + 18x - 16x + 24\) 8 terms with correct multiplication of their 4 terms by \(5x\) | M0 M1A0 |
| Alt 2 \(10x^2 - 19x - 5 \qquad\) implied 4 terms with 3 correct \((3x + 4)(10x^2 - 19x - 5) = 30x^3 - 54x^2 - 15x + 40x^2 - 76x - 20\) 6 terms with correct multiplication of their 3 terms by 4 | M1 M1A0 |
| 1st M1 with a 4-term expansion followed by incorrect simplification to 3 terms can still score the 2nd M1 using their 3 terms | |
| One single expansion is full marks or zero |
11 Expand and simplify fully \(\qquad (3 + 2x)^5\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Evidence of 1 5 10 10 5 1 used for all six coefficients (terms could be written incorrectly) | M1 | the 1s can be ignored but 5 10 10 5 must be seen and used (don’t accept it just being written in Pascal’s triangle) |
| \((3)^5 + 5(3)^4(2x) + 10(3)^3(2x)^2 + 10(3)^2(2x)^3 + 5(3)(2x)^4 + (2x)^5\) | M1dep | oe eg \((3)^5(2x)^0\) written for first term at least 4 terms correct (could already be simplified and missing brackets recovered) |
| \((3)^5 + 5(3)^4(2x) + 10(3)^3(2x)^2 + 10(3)^2(2x)^3 + 5(3)(2x)^4 + (2x)^5\) | M1dep | oe eg \((3)^5(2x)^0\) written for first term all correct |
| \(243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5\) | A1 | |
| Alternative method 2 | ||
| \((3 + 2x)^2 = 9 + 12x + 4x^2\) | M1 | |
| \((3 + 2x)^3 = 27 + 54x + 36x^2 + 8x^3\) | M1dep | oe the terms may not have been collected could do \((3 + 2x)^2 \times (3 + 2x)^2\). If they use this method (doesn’t refer to \((3 + 2x)^3\)) then award this mark for answer expanded correctly but with one numerical error. Terms must be collected |
| \((3 + 2x)^4 = 81 + 216x + 216x^2 + 96x^3 + 16x^4\) | M1dep | terms must be collected could do \((3 + 2x)^2 \times (3 + 2x)^3\). If they use this method (doesn’t refer to \((3 + 2x)^4\)) then award this mark for answer expanded correctly but with one numerical error. Terms must be collected would imply first 2 M marks if done correctly |
| \(243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5\) | A1 | |
| Alternative method 3 | ||
| Evidence of 1 5 10 10 5 1 used for all six coefficients (could be written incorrectly) | M1 | the 1s can be ignored but 5 10 10 5 must be seen and used |
| \(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\) | M1dep | from using a general expansion of \((a + b)^5\) |
| \((3)^5 + 5(3)^4(2x) + 10(3)^3(2x)^2 + 10(3)^2(2x)^3 + 5(3)(2x)^4 + (2x)^5\) | M1dep | oe all correct |
| \(243 + 810x + 1080x^2 + 720x^3 + 240x^4 + 32x^5\) | A1 | |
| Working could be seen as a list or a grid. This can be awarded full marks if done correctly Candidates could use a combination of methods. Use whichever alt method works best (probably alt 2) Missing brackets must be recovered | M3A1 |
6 \(\quad y = 2x^4\left(x^3 + 2 - \dfrac{3}{x}\right)\)
Work out \(\quad \dfrac{\mathrm{d}y}{\mathrm{d}x}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (\(y =\)) \(2x^7 + 4x^4 - 6x^3\) | M1 | for any 2 terms correct |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 14x^6 + 16x^3 - 18x^2\) | A2 | oe eg \(2x^2(7x^4 + 8x - 9)\) A1 for any correct term correctly differentiated |
| Alternative method 2 (product rule) | ||
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\) \(8x^3\left(x^3 + 2 - \dfrac{3}{x}\right) + 2x^4(3x^2 + 3x^{-2})\) | M1 | for either \(2x^4\) differentiated correctly multiplied by the bracket or the bracket differentiated correctly multiplied by \(2x^4\) eg \(8x^3\left(x^3 + 2 - \dfrac{3}{x}\right)\) |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 14x^6 + 16x^3 - 18x^2\) | A2 | oe eg \(2x^2(7x^4 + 8x - 9)\) A1 for any term correct |
| Ignore subsequent incorrect factorisation Condone incorrect use of \(y =\) on the answer line |
6 Prove algebraically that when \(n\) is an integer
\[\frac{(2n + 1)^2 - (2n - 1)^2}{4} \quad \text{is always even.}\][3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Expands the given brackets | ||
| \(((2n + 1)^2 =)\; 4n^2 + 2n + 2n + 1\) or \(((2n - 1)^2 =)\; 4n^2 - 2n - 2n + 1\) | M1 | oe expansion eg \(((2n + 1)^2 =)\; 4n^2 + 4n + 1\) may be seen in a grid may be seen embedded in second mark ignore any denominator |
| \(4n^2 + 2n + 2n + 1 - 4n^2 + 2n + 2n - 1\) or \(4n^2 + 4n + 1 - 4n^2 + 4n - 1\) or \(4n^2 + 2n + 2n + 1 - (4n^2 - 2n - 2n + 1)\) and \(8n\) with no errors seen or \(4n^2 + 4n + 1 - (4n^2 - 4n + 1)\) and \(8n\) with no errors seen | M1dep | terms in any order ignore any denominator |
| M2 seen and valid explanation | A1 | eg1 M2 seen and \(\dfrac{8n}{4} = 2n\) eg2 M2 seen and \(8n\) is even and when divided by 4 it is even |
| Alternative method 2 Difference of two squares | ||
| \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1))\) or \((2n + 1 + 2n - 1)(2n + 1 - 2n + 1)\) | M1 | ignore any denominator |
| M1 seen and \(4n \times 2\) with no errors seen | M1dep | ignore any denominator |
| M2 seen and valid explanation | A1 | eg1 M2 seen and \(\dfrac{4n \times 2}{4} = 2n\) eg2 M2 seen and \(\dfrac{8n}{4} = 2n\) eg3 M2 seen and \(8n\) is even and when divided by 4 it is even |
| Do not allow missing brackets even if recovered | |
| Alt 1 \(4n^2 + 4n + 1 - 4n^2 - 4n + 1\) | M1M0 |
| Alt 1 \(4n^2 + 2n + 2n + 1 - (4n^2 - 2n - 2n + 1)\) \(= 4n^2 + 4n + 1 - 4n^2 - 4n - 1 = 8n\) (\(8n\) but error seen) | M1 M0 |
| Alt 1 Only \(8n\) | M0M0 |
| Alt 1 2nd M1 Allow unnecessary brackets eg \(4n^2 + 4n + 1 - (4n^2 - 4n + 1) = (4n^2 - 4n^2) + (4n + 4n) + (1 - 1)\) | M1M1 |
| Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - 2n - 1)\) | M0M0 |
| Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1))\) \(= (2n + 1 + 2n - 1)(2n + 1 - 2n - 1) = 4n \times 2\) (\(4n \times 2\) but error seen) | M1 M0 |
| Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1)) = 8n\) | M1M0 |
| Alt 2 Only \(4n \times 2\) | M0M0 |
| Response only referring to odds and evens or only involving substitution | M0M0A0 |
| Assuming the expression simplifies to \(2k\) and working back could score up to M1M1 | |
| Setting up an equation eg \((2n + 1)^2 - (2n - 1)^2 = 4\) could score up to M1M1 | |
| For A1 do not allow incorrect use of = eg \(4n^2 + 2n + 2n + 1 - 4n^2 + 2n + 2n - 1\) \(= \dfrac{8n}{4} = 2n\) | M1M1 A0 |
5 \(\quad (3x + a)(5x - 4) \equiv 15x^2 - 2x + b\)
Work out the values of \(a\) and \(b\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(15x^2 - 12x + 5ax - 4a\) or \(5ax - 12x = -2x\) or \(5a - 12 = -2\) or \(b = -4a\) | M1 | oe |
| (\(a =\)) 2 | A1 | |
| (\(b =\)) \(-4 \times\) their \(a\) | A1ft | \(-8\), but do not award \(-8\) unless it comes from \(a = 2\) |
| Candidates who use substitutions for \(x\) are likely to use \(x = 0\) and gain M1. Award M1 for any number substituted in correctly to gain an equation in \(a\) and \(b\) |
11 Expand and simplify fully \(\quad (5x + 3y^2)(4x - y^2)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| 3 terms from \(20x^2 \quad -5xy^2 \quad (+)12xy^2 \quad -3y^4\) | M1 | may be seen in a grid |
| \(20x^2 - 5xy^2 + 12xy^2 - 3y^4\) | A1 | four correct terms in any order may be seen in a grid implied by correct answer |
| \(20x^2 + 7xy^2 - 3y^4\) | A1 | terms may be in any order |
| Terms seen in a grid must have the correct signs | |
| Terms must be fully processed eg do not allow \(4x3y^2\) unless recovered | |
| \(xy^2\) may be \(y^2x\) throughout | |
| \(20x^2 + 7xy^2 - 3y^4\) followed by incorrect further work | M1A1A0 |
9 Expand and simplify fully \(\quad (2x - 5)(3x - 4)(x + 2)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Expanding second and third bracket \(3x^2 + 6x - 4x - 8\) with at least three terms correct or \(3x^2 + 2x + k\) where \(k\) is a non-zero constant | M1 | oe expansion attempt of one pair of brackets eg1 first and second brackets \(6x^2 - 8x - 15x + 20\) with at least three terms correct or \(6x^2 - 23x + k\) where \(k\) is a non-zero constant eg2 first and third brackets \(2x^2 + 4x - 5x - 10\) with at least three terms correct or \((2x^2 - x + k)\) where \(k\) is a non-zero constant |
| \(6x^3 + 12x^2 - 8x^2 - 16x\) or \(6x^3 + 4x^2 - 16x\) or \(-15x^2 - 30x + 20x + 40\) or \(-15x^2 - 10x + 40\) | M1dep | attempt at a full expansion with correct multiplication of their 3 or 4 terms by one of the terms in the remaining bracket oe eg1 \(6x^3 - 8x^2 - 15x^2 + 20x\) or \(6x^3 - 23x^2 + 20x\) or \(12x^2 - 16x - 30x + 40\) or \(12x^2 - 46x + 40\) eg2 \(6x^3 + 12x^2 - 15x^2 - 30x\) or \(6x^3 - 3x^2 - 30x\) or \(-8x^2 - 16x + 20x + 40\) or \(-8x^2 + 4x + 40\) |
| \(6x^3 - 11x^2 - 26x + 40\) | A1 | fully correct simplified expansion allow terms written in a different order |
| For the M marks terms may be seen in a grid | |
| First M1 do not allow any omissions or extra terms eg1 \(3x^2 + 6x - 8\) eg2 \(3x^2 + 6x - 4x - 8 + x^2\) | M0 M0 |
7 Solve \(\quad 2x^2 + 4 \gt (2x - 3)(x + 1)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2x^2 - x - 3\) or \(2x^2 - 3x + 2x - 3\) | M1 | |
| \(4 \gt -x - 3\) | M1dep | oe eg \(7 \gt -x\) |
| \(x \gt -7\) or \(-7 \lt x\) | A1 |
| \(=\) used instead of \(\gt\) throughout and not recovered on answer line | M2A0 |
1 Expand and simplify \(\quad 5(2x - 1) + 4(11 - x)\)
Give your answer in the form \(\quad a(bx + c) \quad\) where \(a\), \(b\) and \(c\) are integers greater than 1 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(10x - 5\) or \(44 - 4x\) | M1 | may be seen in a grid |
| \(10x - 5 + 44 - 4x\) or \(6x + 39\) or \(3 \times (2x + 13)\) | M1dep | oe all terms correct in a single expression |
| \(3(2x + 13)\) | A1 |
21 Show that \(\quad (2n + 3)^3 + n^3 \quad\) is divisible by 9 for all integer values of \(n\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4n^2 + 6n + 6n + 9\) or \(4n^2 + 12n + 9\) | M1 | allow one error implied by \(4n^2 + 12n + k\) or \(an^2 + 12n + 9\) |
| \(8n^3 + 12n^2 + 24n^2 + 36n + 18n + 27\) | M1dep | oe ft their \(4n^2 + 6n + 6n + 9\) allow one error |
| \(8n^3 + 36n^2 + 54n + 27\) or \(9n^3 + 36n^2 + 54n + 27\) | A1 | |
| \(9n^3 + 36n^2 + 54n + 27\) and \(9(n^3 + 4n^2 + 6n + 3)\) | A1 | oe eg \((9n^3 + 36n^2 + 54n + 27) \div 9\) \(= n^3 + 4n^2 + 6n + 3\) or \(9n^3 + 36n^2 + 54n + 27\) and all coefficients are divisible by 9 |
12 This shape is made from two rectangles.
All dimensions are in centimetres.
Not drawn accurately
Show that \(\quad y = 126 - 45x\) [2 marks]
Show that \(\quad A = 2520x - 450x^2\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(30x + 20x + 15x + 10x + 15x + y + y = 252\) or \(90x + 2y = 252\) | M1 | oe |
| \(y = \dfrac{252 - 90x}{2}\) and \(y = 126 - 45x\) | A1 | must see working for M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(30x \times 15x + 20x \times (126 - 45x)\) or \(15x \times 10x + 20x \times (126 - 45x + 15x)\) or \(15x \times 10x + 20x \times (126 - 30x)\) | M1 | oe |
| \(450x^2 + 2520x - 900x^2 = 2520x - 450x^2\) or \(150x^2 + 2520x - 900x^2 + 300x^2\) \(= 2520x - 450x^2\) or \(150x^2 + 2520x - 600x^2 = 2520x - 450x^2\) | A1 | must see correct expansion of brackets |
| Answer | Mark | Comments |
|---|---|---|
| \(2520 - 900x\) | M1 | |
| their \((2520 - 900x) = 0\) or \(x = 2.8\) | M1dep | oe |
| 3528 | A1 |
3
\[3a(2x - 1) + 4(ax + 5) \equiv 60x + b\]Work out the values of \(a\) and \(b\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(6ax - 3a + 4ax + 20\) | M1 | both brackets expanded correctly |
| \(6a + 4a = 60\) or \(-3a + 20 = b\) | M1 | either coefficient equated correctly |
| \(a = 6\) | A1 | |
| \(b = 2\) | A1 | |
| Alternative method 2 | ||
| Correct substitutions leading to two correct equations in \(a\) and \(b\) | M1 | |
| A correct attempt to eliminate either \(a\) or \(b\) | M1 | any valid method |
| \(a = 6\) | A1 | |
| \(b = 2\) | A1 | |
17 Show that \(\quad (x + 1)(x + 3)(x + 4) - x(x^2 + 7x + 11)\)
can be written in the form \(\quad (x + a)(x + b) \quad\) where \(a\) and \(b\) are positive integers. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + 3x + x + 3\) with three terms correct or \(x^2 + 4x + k\) where \(k\) is a non-zero constant | M1 | oe expansion attempt of one pair of brackets eg1 \(\ x^2 + 4x + 3x + 12\) with three terms correct or \(x^2 + 7x + k\) where \(k\) is a non-zero constant eg2 \(\ x^2 + 4x + x + 4\) with three terms correct or \(x^2 + 5x + k\) where \(k\) is a non-zero constant |
| \(x^3 + 3x^2 + x^2 + 3x\) or \(x^3 + 4x^2 + 3x\) or \(4x^2 + 12x + 4x + 12\) or \(4x^2 + 16x + 12\) | M1dep | attempt at a full expansion with correct multiplication of their 3 or 4 terms by one of the terms in the remaining bracket oe eg \(x^3 + 4x^2 + 3x^2 + 12x\) or \(x^3 + 7x^2 + 12x\) or \(x^2 + 4x + 3x + 12\) or \(x^2 + 7x + 12\) (\(x^2 + 7x + 12\) must be from an attempt at a full expansion) or \(x^3 + 4x^2 + x^2 + 4x\) or \(x^3 + 5x^2 + 4x\) or \(3x^2 + 12x + 3x + 12\) or \(3x^2 + 15x + 12\) |
| \(x^3 + 8x^2 + 19x + 12\) | A1 | fully correct expansion allow if terms not collected eg \(x^3 + 3x^2 + x^2 + 3x + 4x^2 + 12x + 4x + 12\) or \(x^3 + 4x^2 + 3x + 4x^2 + 16x + 12\) |
| \(x^2 + 8x + 12\) | A1ft | ft M2A0 full simplification of their \((x^3 + 8x^2 + 19x + 12) - x^3 - 7x^2 - 11x\) their \((x^3 + 8x^2 + 19x + 12)\) must be a cubic |
| \(x^2 + 8x + 12\) and \((x + 6)(x + 2)\) or \((x + 2)(x + 6)\) | A1 | oe product of brackets |
| 1st M1 Do not allow omissions or extras eg1 \(\ x^2 + 3x + 3\) eg2 \(\ x^2 + 3x + x + 3 + x^2\) | M0 M0 |
| For the first 2 marks terms may be seen in a grid | |
| If 1st A1 has been awarded with terms not collected, A1ft can still be awarded using their simplified cubic eg \(\ x^3 + 4x^2 + 3x + 4x^2 + 16x + 12\) \(= x^3 + 8x^2 + 18x + 12\) \(x^3 + 8x^2 + 18x + 12 - x^3 - 7x^2 - 11x\) \(= x^2 + 7x + 12\) | M1M1A1 A1ftA0 |
| First A1 may be seen embedded eg \(\ x^3 + 8x^2 + 19x + 12 - x^3 - 7x^2 - 11x\) | M1M1A1 |
| If an attempt at the expansion of all three brackets in one go is made it must be fully correct to gain M2A1, otherwise M0M0A0 eg \(\ x^2 + 3x + x + 3 + x^2 + 4x\) | M0M0A0 |
| Allow recovery of missing brackets when subtracting \(x^3 + 7x^2 + 11x\) from their cubic | |
| For final A1 allow \(x^2 + 8x + 12\) and \(a = 6 \ \ b = 2\) or \(x^2 + 8x + 12\) and \(a = 2 \ \ b = 6\) | |
| Ignore equating to zero and/or any ‘solving’ of an equation |
16 \(A = 2 - 5x \qquad B = 3x - 1 \qquad C = x^2\)
Show that \(\qquad (2A + 3B)^2 \equiv A + B + C\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\) | M1 | |
| \(4 - 10x + 9x - 3 = 1 - x\) | M1dep | |
| \((1 - x)^2 = 1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
| Alternative method 2 | ||
| \(4(2 - 5x)^2 + 6(2 - 5x)(3x - 1)\) \(+\ 6(2 - 5x)(3x - 1) + 9(3x - 1)^2\) | M1 | oe allow \(+\ 12(2 - 5x)(3x - 1)\) for \(+\ 6(2 - 5x)(3x - 1) + 6(2 - 5x)(3x - 1)\) |
| \(4(4 - 10x - 10x + 25x^2)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 9(9x^2 - 3x - 3x + 1)\) \(= 16 - 40x - 40x + 100x^2 + 36x - 12\) \(-\ 90x^2 + 30x + 36x - 12 - 90x^2\) \(+\ 30x + 81x^2 - 27x - 27x + 9\) | M1dep | oe must see expansions must see working for 1st M1 allow \(+\ 12(6x - 2 - 15x^2 + 5x)\) for \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) |
| \(1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
| Alternative method 3 | ||
| \(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\) | M1 | oe |
| \((4 - 10x + 9x - 3)^2\) \(= 16 - 40x + 36x - 12 - 40x + 100x^2\) \(-\ 90x^2 + 30x + 36x - 90x^2 + 81x^2\) \(-\ 27x - 12 + 30x - 27x + 9\) | M1dep | oe must see expansions |
| \(1 - 2x + x^2\) | A1 | must see working for M2 |
| \(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\) | B1 | |
| Allow working down both sides of an equation/identity | |
| M2A1 is for working on \((2A + 3B)^2\) | |
| B1 is for working on \(A + B + C\) | |
| \(1 - 2x + x^2\) with working for M2 seen and \(2 - 5x + 3x - 1 + x^2 = x^2 - 2x + 1\) | 4 marks |
| \(1 - x^2 = 1 - 2x + x^2\) (do not allow missing brackets even if recovered) |
15 Work out the rate of change of \(y\) with respect to \(x\) at the point on the curve
\(y = x^2(x^2 - 9) \qquad\) where \(\ x = -2\)
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^4 - 9x^2\) | M1 | |
| \(4x^3\) or \((-)2 \times 9x\) or \((-)18x\) | M1 | differentiates at least one of their terms their term must be a function of \(x\) after an attempt to expand brackets \(4x^3 - 18x\) implies M2 |
| \(4 \times (-2)^3 - 18 \times -2\) or \(4 \times -8 - 18 \times -2\) or \(-32 + 36\) | M1dep | oe dep on 2nd M1 substitutes \(-2\) into their \(4x^3 - 18x\) their \(4x^3 - 18x\) must be two terms, each a function of \(x\) |
| 4 | A1 |
| Allow recovery of brackets | |
| \(4 \times -2^3\) is allowed for \(\ 4 \times (-2)^3\) | |
| 3rd M can still be awarded even if further substitution seen eg1 \(\ 4x^3 - 18x \quad 4 \times (-2)^3 - 18 \times -2\) and \(4^4 - 9 \times 4^2 = 112\) Answer 112 eg2 \(\ 4x^3 - 18x \quad 4 \times (-2)^3 - 18 \times -2 = 4\) and \(4 \times 4^3 - 18 \times 4 = 184\) Answer 184 | M3 A0 M3 A0 |
| Only substituting \(x = -2\) into second derivative can score a maximum of M1M1M0A0 | |
| 4 followed by answer \(y = 4\) | M3A1 |
| 4 followed by answer \(y = 4x\) | M3A0 |
| Do not regard substitution of \(x = 2\) as a misread | |
| Beware that finding the gradient of the line through \((-2,\ -20)\) and \((3,\ 0)\) gives answer 4 | Zero |
| Beware that \((-)18x\) may come from wrong method eg1 \(\ 2x(2x - 9) = 4x^2 - 18x\) eg2 \(\ 2x(x^2 - 9) = 2x^3 - 18x\) | Zero Zero |
14 Work out the value of \(\quad \left(3^{\frac{1}{2}} + 3^{\frac{3}{2}}\right)^2\)
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(3^{\frac{1}{2}} \times 3^{\frac{1}{2}} + 3^{\frac{1}{2}} \times 3^{\frac{3}{2}} + 3^{\frac{1}{2}} \times 3^{\frac{3}{2}} + 3^{\frac{3}{2}} \times 3^{\frac{3}{2}}\) or \(\sqrt{3}\sqrt{3} + \sqrt{3}\sqrt{27} + \sqrt{3}\sqrt{27} + \sqrt{27}\sqrt{27}\) | M1 | oe allow an error in one term |
| 3 or 9 or 27 | M1dep | |
| 48 | A1 | |
| Alternative method 2 | ||
| \(\sqrt{3}\) and \(3\sqrt{3}\) | M1 | \(3\sqrt{3}\) must come from correct working |
| \((4\sqrt{3})^2\) | M1dep | |
| 48 | A1 | |
| Alternative method 3 | ||
| \(\left(3^{\frac{1}{2}}\right)^2(1 + 3)^2\) | M1 | oe |
| \(3 \times 4^2\) | M1dep | oe |
| 48 | A1 | |
| Alt 1 mark scheme … likely to see a 3 (or 9 or 27) somewhere, so need to be careful that the M1 mark has been earned before awarding A1 | |
| In alt 1, for the first M1, we want to see an attempt at the full expansion of the correct terms. Probably 4 terms, but there could be 3 if they combine the middle two terms. eg \((\sqrt{3} + 27)(\sqrt{3} + 27)\) scores M0 because it ought to be \(\sqrt{27}\) not 27 |
11 Expand and simplify fully \(\qquad (x + 2)(x + 3)(x + 4)\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 expands \((x + 2)(x + 3)\) first | ||
| \(x^2 + 3x + 2x + 6\) or \(x^2 + 5x + 6\) | M1 | oe must have a term in \(x^2\) allow one error but no omissions or extras implied by \(x^2 + 5x + k\) or \(ax^2 + 5x + 6\) |
| \(x^3 + 5x^2 + 6x + 4x^2 + 20x + 24\) | M1dep | oe eg \(x^3 + 3x^2 + 2x^2 + 6x + 4x^2 + 12x + 8x + 24\) allow one further error but no omissions or extras |
| \(x^3 + 9x^2 + 26x + 24\) | A1 | |
| Alternative method 2 expands \((x + 3)(x + 4)\) first | ||
| \(x^2 + 3x + 4x + 12\) or \(x^2 + 7x + 12\) | M1 | oe must have a term in \(x^2\) allow one error but no omissions or extras implied by \(x^2 + 7x + k\) or \(ax^2 + 7x + 12\) |
| \(x^3 + 7x^2 + 12x + 2x^2 + 14x + 24\) | M1dep | oe eg \(x^3 + 3x^2 + 4x^2 + 12x + 2x^2 + 6x + 8x + 24\) allow one further error but no omissions or extras |
| \(x^3 + 9x^2 + 26x + 24\) | A1 | |
| Alternative method 3 expands \((x + 2)(x + 4)\) first | ||
| \(x^2 + 4x + 2x + 8\) or \(x^2 + 6x + 8\) | M1 | oe must have a term in \(x^2\) allow one error but no omissions or extras implied by \(x^2 + 6x + k\) or \(ax^2 + 6x + 8\) |
| \(x^3 + 6x^2 + 8x + 3x^2 + 18x + 24\) | M1dep | oe eg \(x^3 + 4x^2 + 2x^2 + 8x + 3x^2 + 12x + 6x + 24\) allow one further error but no omissions or extras |
| \(x^3 + 9x^2 + 26x + 24\) | A1 | |
| For M marks terms may be seen in a grid (+ signs not needed) | |
| Correct answer followed by further work | M2A0 |
| Ignore further simplification after 4 terms seen eg Alt 1 \(\ x^2 + 3x + 2x + 6 = x^2 + 6x + 6\) \((x^2 + 6x + 6)(x + 4) \rightarrow x^3 + 4x^2 + 6x^2 + 24x + 6x + 18\) (error) | M1 M1depA0 |
| Second M1 Must be the product of a two term bracket and a three or four term bracket | |
| Missing brackets may be recovered |
5 In the expansion and simplification of \(\quad (x - 3)(x^2 + 5x + k) \quad\) the coefficient of \(x^2\) is equal to the coefficient of \(x\).
\(k\) is a constant.
Work out the value of \(k\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x^3) + 5x^2 + kx - 3x^2 - 15x\ (-\ 3k)\) | M1 | allow one sign error in the \(x^2\) or \(x\) terms |
| their \((5 - 3)\) = their \((k - 15)\) | M1dep | \(5x^2 - 3x^2 = kx - 15x\) on its own, is not enough for M1dep |
| 17 | A1 |
| For the first M1, we do not need to see the \(x^3\) term or the \(-\ 3k\) term, but we do need to see the other 4 terms (3 terms, if they combine the \(x^2\) terms). The terms of the expansion might appear in a grid, which can score the first M1 Mark positively … terms in a grid might differ from terms written as a string of terms … mark the better version. |
12
| Answer | Mark | Comments |
|---|---|---|
| \(3(25 - x^2)\) or \(-3(x^2 - 25)\) or \((15 + 3x)(5 - x)\) or \((x + 5)(15 - 3x)\) | M1 | oe partial factorisation eg \(-(3x + 15)(x - 5)\) Brackets in either order Do not allow \(-(3x^2 - 75)\) |
| \(3(5 + x)(5 - x)\) or \(3(-x - 5)(x - 5)\) or \(-3(x + 5)(x - 5)\) or \(-3(5 - x)(-x - 5)\) | A1 |
| \((-x + 5)\) is equivalent to \((5 - x)\) etc | |
| Do not allow A1 for incorrect notation in final answer eg \((5 + x)3(5 - x)\) | M1A0 |
| Do not allow A1 for use of multiplication signs in final answer eg \(3 \times (5 + x) \times (5 - x)\) | M1A0 |
| Correct answer followed by incorrect further work | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(9n^2 + 3n + 3n + 1\) or \(9n^2 + 6n + 1\) or \(9n^2 - 3n - 3n + 1\) or \(9n^2 - 6n + 1\) | M1 | oe Terms may be seen in a grid |
| \(12n\) with no incorrect working | A1 | Brackets can be recovered |
| Alternative method 2 | ||
| \((3n + 1 + 3n - 1)(3n + 1 - (3n - 1))\) or \((3n + 1 + 3n - 1)(3n + 1 - 3n + 1)\) | M1 | oe Brackets around \(3n - 1\) can be recovered |
| \(12n\) | A1 | |
| Alt 1 \(12n\) may come from incorrect working eg1 \(3n^2 + 6n + 1 - (3n^2 - 6n + 1) = 12n\) eg2 \(9n^2 + 3n + 1 - (9n^2 - 3n + 1) = 12n\) | M0A0 M0A0 |
| Alt 1 Recovery of brackets eg1 \(9n^2 + 6n + 1 - 9n^2 - 6n + 1 = 12n\) eg2 \(9n^2 + 6n + 1 - 9n^2 - 6n + 1 = 2\) | M1A1 M1A0 |
| Alt 2 Recovery of brackets eg1 \((3n + 1 + 3n - 1)(3n + 1 - 3n - 1) = 12n\) eg2 \((3n + 1 + 3n - 1)(3n + 1 - 3n - 1) = 0\) | M1A1 M0A0 |
| Do not allow A1 for use of multiplication signs in final answer eg \(12 \times n\) with no incorrect working | M1A0 |
5 \((x + 4)(x^2 - kx - 5) \quad\) is expanded and simplified.
The coefficient of the \(x^2\) term is twice the coefficient of the \(x\) term.
Work out the value of \(k\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(x^3\) +) \(4x^2 - kx^2 - 4kx - 5x\) (\(-20\)) | M1 | or \(4 - k\) and \(-4k - 5\) seen as coefficients |
| \(4 - k = 2(-4k - 5)\) | M1dep | ft their expansion if first M mark earned |
| (\(k\) =) \(-2\) | A1 |
Condone one sign error in the first two steps
Ignore errors in \(x^3\) and \(-20\) for the first M1
11 The \(x^2\) term in the expansion of \(\quad (3x + 4)(x^2 + px + 5) \quad\) is \(\ -23x^2\)
Work out the value of \(p\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(4x^2\) or \(3px^2\) or \(4 + 3p\) | M1 | May be seen in an expansion or a grid Allow unsimplified eg \(3x \times px\) |
| their \(4(x^2)\) + their \(3p(x^2) = -23(x^2)\) | M1dep | Correct or ft their expansion ft is equating their terms in \(x^2\) to \(-23x^2\) Must be at least two terms with at least one linear term in \(p\) Allow unsimplified eg \(3x \times px + 4x^2 = -23x^2\) |
| \(-9\) | A1 |
| In this question, only consider terms in \(x^2\) | |
| If only one term in \(x^2\) the maximum mark is M1 | |
| \(4 + 3p = -23\) followed by \(7p = -23\) | M1 M1 A0 |
9 Expand and simplify \(\qquad (x - 5)^3\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(x^2 - 5x - 5x + 25 \quad\) or \(\quad x^2 - 10x + 25\) | M1 | allow one error |
| \(x^3 - 5x^2 - 5x^2 + 25x - 5x^2 + 25x + 25x - 125\) or \(x^3 - 10x^2 + 25x - 5x^2 + 50x - 125\) | M1dep | oe ft their \(x^2 - 5x - 5x + 25 \quad\) or \(\quad x^2 - 10x + 25\) and allow one error only if no errors made for the 1st M1 |
| \(x^3 - 15x^2 + 75x - 125\) | A1 | |
| Alternative method 2 | ||
| \(1x^3 + 3x^2(-5) + 3x(-5)^2 + 1(-5)^3\) | M1 | Using 1 3 3 1 ... coefficients from Pascal's triangle for the cubic expansion |
| \(x^3 - 15x^2 + 75x - 125\) | M1 | allow one error |
| \(x^3 - 15x^2 + 75x - 125\) | A1 | |
Penalise further work
There must be three or four terms for the 1st M1 in alt 1
In alt 1 ... for M1 M1 they must make at most one error in the 1st two steps
In the 2nd step, \(-50x\) (instead of \(+50x\)) and \(-10x\) (instead of \(+50x\)) are examples of one error
In alt 2 ...
\(1x^3 + 3x^2(-5) + 3x(-5)^2 + 1(-5)^3 = x^3 - 15x^2 + 75x - 125\) written directly, with no intermediate steps needed or seen, automatically scores 3 marks
No questions match these filters.