June 2016 Paper 1 Q14
14
(a) \((x - 3)\) is a factor of \(\qquad x^3 - 8x^2 + ax + 42 \qquad\) where \(a\) is an integer.
Show that the value of \(a\) is 1 [2 marks]
(b) Hence, factorise fully \(\qquad x^3 - 8x^2 + x + 42\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((3)^3 - 8(3)^2 + 3a + 42 = 0\) or \(27 - 72 + 3a + 42 = 0\) | M1 | Equating to zero might not be seen until later in the working. |
| \(3a = 3\) | A1 | \(3a = 3\) implies \(3a - 3 = 0\) |
| Alternative method 2 | ||
| \((x^3 - 8x^2 + ax + 42) \div (x - 3)\) to give a quotient of \(x^2 - 5x + (a - 15)\) and a remainder of \(3a - 3\) | M1 | |
| Remainder = 0 so \(3a = 3\) | A1 | |
| Alternative method 3 | ||
| \(x^3 - 8x^2 + ax + 42\) \(= (x - 3)(x^2 + px - 14)\) Comparing \(x^2\) coefficients gives \(p = -5\) | M1 | |
| Using \(p = -5\) and comparing \(x\) coefficients gives \(a = 1\) | A1 | |
Additional guidance
| In alt 1 ... assuming that \(a = 1\) and showing that substituting \(x = 3\) in the expression gives zero is only verifying the result ... and scores SC1 Similarly, assuming \(a = 1\) and working as in alt 2 and alt 3 to verify the result. | SC1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(x^3 - 8x^2 + x + 42\) \(\equiv (x - 3)(x^2 + kx - 14)\) | M1 | Sight of quadratic with \(x^2\) and \(-14\) as the first and last terms |
| \((x + 2)\) or \((x - 7)\) | A1 | |
| \((x - 3)(x + 2)(x - 7)\) | A1 | any order |
| Alternative method 2 | ||
| Substitutes another value into the expression and tests for ‘= 0’ | M1 | their value correctly substituted eg. \(2^3 - 8(2)^2 + 2 + 42\ (= 20) \neq 0\) |
| \((x + 2)\) or \((x - 7)\) | A1 | |
| \((x - 3)(x + 2)(x - 7)\) | A1 | any order |
| Alternative method 3 | ||
| Long division of polynomials getting as far as \(x^2 - 5x\ \ldots\ldots\) | M1 | \((x^3 - 8x^2 + x + 42) \div (x - 3) = x^2 - 5x - 14\) |
| \((x + 2)\) or \((x - 7)\) | A1 | |
| \((x - 3)(x + 2)(x - 7)\) | A1 | any order |
Additional guidance
An answer of \((x + 2)(x - 7)\) ie \((x - 3)\) missing ... implies M1 A1
An answer of \((x - 3)(x - 2)(x + 7)\) scores SC1 ... sign errors in two factors
Ignore 'solutions' ie \(x = 3\), \(-2\) and 7