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More questions on this topic: Level 2 FM Algebra: Polynomials (10) AS Pure: Polynomials (71) A2 Pure: Polynomials (39)
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 |
19 \(\quad 8x^2 + 20x + n \equiv c(x + d)^2 + 3 \quad\) where \(c\), \(d\) and \(n\) are constants.
Work out the values of \(c\), \(d\) and \(n\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(c =\)) 8, (\(d =\)) \(\dfrac{5}{4}\) or 1.25 and (\(n =\)) \(\dfrac{31}{2}\) or \(15\dfrac{1}{2}\) or 15.5 | B3 | oe B2 any 2 correct on answer line (not embedded) B1 (\(d =\)) \(\dfrac{5}{4}\) or 1.25 or (\(n =\)) \(\dfrac{31}{2}\) or \(15\dfrac{1}{2}\) or 15.5 on answer line or \(cx^2 + 2cdx + cd^2\) (+3) or \(8\left(x + \dfrac{5}{4}\right)^2\) or \(8(x + 1.25)^2\) in method |
| (\(c =\)) 8 on its own is 0 marks | M0 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 \(b\) |
21 \(3x^2 + 2bx + 8a \quad\) can be written in the form \(\quad 3(x + a)^2 + b + 2\)
Work out the two possible pairs of values of \(a\) and \(b\). [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(3(x^2 + ax + ax + a^2) \ldots\) or \(3(x^2 + 2ax + a^2) \ldots\) or \(3\left(x + \dfrac{b}{3}\right)^2 \ldots\) or \(2b = 6a\) or \(8a = 3a^2 + b + 2\) | M1 | oe eg \(3x^2 + 6ax + 3a^2 \ldots\) or \(\dfrac{b}{3} = a\) or \(b + 2 = -3\left(\dfrac{b}{3}\right)^2 + 8a\) |
| \(2b = 6a\) and \(8a = 3a^2 + b + 2\) | M1dep | oe equations eg \(\dfrac{b}{3} = a\) and \(b + 2 = -3\left(\dfrac{b}{3}\right)^2 + 8a\) |
| \(3a^2 + 3a - 8a + 2\ (= 0)\) or \(3a^2 - 5a + 2\ (= 0)\) | M1dep | oe quadratic equation in \(a\) |
| \((3a - 2)(a - 1)\) or \(\dfrac{--5 \pm \sqrt{(-5)^2 - 4 \times 3 \times 2}}{2 \times 3}\) | M1 | oe eg \(\dfrac{5}{6} \pm \sqrt{\dfrac{25}{36} - \dfrac{2}{3}}\) ft their 3-term quadratic |
| \(a = \dfrac{2}{3}\) and \(a = 1\) or \(a = \dfrac{2}{3}\) and \(b = 2\) or \(a = 1\) and \(b = 3\) | A1 | |
| \(a = \dfrac{2}{3}\) and \(b = 2\) and \(a = 1\) and \(b = 3\) | A1 | |
| Alternative method 2 | ||
| \(3(x^2 + ax + ax + a^2) \ldots\) or \(3(x^2 + 2ax + a^2) \ldots\) or \(3\left(x + \dfrac{b}{3}\right)^2 \ldots\) or \(2b = 6a\) or \(8a = 3a^2 + b + 2\) | M1 | oe eg \(3x^2 + 6ax + 3a^2 \ldots\) or \(\dfrac{b}{3} = a\) or \(b + 2 = -3\left(\dfrac{b}{3}\right)^2 + 8a\) |
| \(2b = 6a\) and \(8a = 3a^2 + b + 2\) | M1dep | oe equations eg \(\dfrac{b}{3} = a\) and \(b + 2 = -3\left(\dfrac{b}{3}\right)^2 + 8a\) |
| \(\dfrac{8b}{3} = 3\left(\dfrac{b}{3}\right)^2 + b + 2\) or \(b^2 - 5b + 6\ (= 0)\) | M1dep | oe quadratic equation in \(b\) |
| \((b - 2)(b - 3)\) or \(\dfrac{--5 \pm \sqrt{(-5)^2 - 4 \times 1 \times 6}}{2 \times 1}\) | M1 | oe eg \(\dfrac{5}{2} \pm \sqrt{\dfrac{25}{4} - 6}\) ft their 3-term quadratic |
| \(b = 2\) and \(b = 3\) or \(a = \dfrac{2}{3}\) and \(b = 2\) or \(a = 1\) and \(b = 3\) | A1 | |
| \(a = \dfrac{2}{3}\) and \(b = 2\) and \(a = 1\) and \(b = 3\) | A1 | |
Allow \(0.\dot{6}\) for \(\dfrac{2}{3}\)
Allow 0.67 for \(\dfrac{2}{3}\) for first A1
In quadratic formula allow \(5^2\) for \((-5)^2\) but use of \(-5^2\) must be recovered
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 | |
4 \(p(x - 1) + 2(3x + k) \equiv 4(x + 2) \qquad\) where \(p\) and \(k\) are integers.
Work out the values of \(p\) and \(k\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(px - p + 6x + 2k = 4x + 8\) or \(px + 6x = 4x\) or \(p + 6 = 4\) | M1 | oe |
| \(p = -2\) | A1 | This could imply first M mark if not seen |
| \(2k - \text{their } p = 8\) or \(2k = \text{their } p + 8\) | M1 | oe could be awarded by substituting a value of \(x\) with \(p = -2\) |
| \(k = 3\) | A1ft | need to check back for ft mark |
| Alternative method 2 | ||
| A correct equation obtained by substituting a value for \(x\) in the identity | M1 | eg \(\ x = 0 \quad 2k - p = 8\) \(\phantom{\text{eg }} x = 1 \quad p - p + 6 + 2k = 12\) \(\phantom{\text{eg }} x = 2 \quad 2p - p + 12 + 2k = 16\) |
| A second correct equation obtained by substituting a value for \(x\) in the identity | M1 | oe could go back to equating coefficients at this stage |
| \(p = -2\) | A1 | |
| \(k = 3\) | A1 | may come from one equation by substituting \(x = 1\) |
| Correct expansion, then \(p + 6 = 4\) followed by \(p = 2\) (incorrect) would give \(k = 5\) on ft … allow ft mark for \(k\) | M1, A0 M1, A1ft |
| In Alt 2 substituting \(x = 1\) leads to \(k = 3\) (a second equation would be needed to gain further marks) | M1, A1 |
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. |
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 |
6 \(mx + 4 - 2(x + p) \equiv 6(x + 1) \qquad\) where \(m\) and \(p\) are integers.
Work out the values of \(m\) and \(p\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(mx + 4 - 2x - 2p\) or \(6x + 6\) | M1 | allow one error on the left hand side |
| \(mx - 2x = 6x\) or \(4 - 2p = 6\) | M1 | oe eg. \(m - 2 = 6\) ft their expansion if 1st M1 earned |
| \(m = 8\) | A1 | |
| \(p = -1\) | A1 | |
| Alternative method 2 | ||
| \(mx + 4 - 2x - 2p = 6x + 6\) | M1 | At most one error in total in the expansion and the simplifying |
| \(mx - 2p = 8x + 2\) | M1 | This must be in the form \(ax + b = cx + d\) |
| \(m = 8\) | A1 | |
| \(p = -1\) | A1 | |
| Alternative method 3 | ||
| An equation obtained by substituting a value for \(x\) in the identity | M1 | eg \(\quad x = 0 \quad 4 - 2p = 6\) \(\qquad x = 1 \quad m + 4 - 2 - 2p = 12\) \(\qquad x = 2 \quad 2m + 4 - 4 - 2p = 18\) |
| A second equation obtained by substituting a value for \(x\) in the identity | M1 | |
| \(m = 8\) | A1 | |
| \(p = -1\) | A1 | |
\(mx + 4 - 2x + 2p\) is one error in the left hand side
\(mx + 4 - 2x - p\) is one error in the left hand side
\(mx + 4 - 2x + p\) is two errors in the left hand side
In alt 2 ... allow at most one error in the expansion and the simplifying (the first two M marks) ... one error can then score M1 M1, two errors will score M1 M0
Also, in alt 3, allow at most one error in forming their two simultaneous equations
No questions match these filters.