June 2018 Paper 1 Q17

AQAOld spec4 marksPolynomials

(a) Use the factor theorem to show that \((x - 2)\) is a factor of \(\quad x^3 + 8x^2 + 5x - 50\) [1 mark]

(b) Hence, factorise fully \(\quad x^3 + 8x^2 + 5x - 50\) [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(\mathrm{f}(2) = (2)^3 + 8(2)^2 + 5(2) - 50\) \(= 8 + 32 + 10 - 50 = 0\)	B1	substitutes \(x = 2\) and verifies that \(\mathrm{f}(2) = 0\) … the terms must be evaluated

Using the factor theorem is essential. Using long division here scores M0

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1
\(x^3 + 8x^2 + 5x - 50\) \(\equiv (x - 2)(x^2 + kx + 25)\)	M1	Sight of a 3 term quadratic with \(x^2\) and \(+25\) as the first and last terms
\(x^2 + 10x + 25\)	A1
\((x - 2)(x + 5)^2\)	A1	oe
Alternative method 2
Substitutes another value into the expression and tests for ‘= 0’	M1	their value correctly worked out eg \(\mathrm{f}(1) = -36 \quad \mathrm{f}(3) = 64\)
\((x + 5)\)	A1	coming from \(\mathrm{f}(-5) = -125 + 200 - 25 - 50 = 0\)
\((x - 2)(x + 5)^2\)	A1	oe
Alternative method 3
Long division of polynomials getting as far as \(x^2 + 10x\) ……	M1
\(x^2 + 10x + 25\)	A1
\((x - 2)(x + 5)^2\)	A1	oe
Alternative method 4
Using synthetic division to arrive at \(x^2 + 10x\) ……	M1	\(\begin{array}{r\|rrrr} 2 & 1 & 8 & 5 & -50 \\ & 0 & 2 & 20 & 50 \\ \hline & 1 & 10 & 25 & 0 \end{array}\)
\(x^2 + 10x + 25\)	A1
\((x - 2)(x + 5)^2\)	A1	oe
Alternative method 5
\(x^3 + 8x^2 + 5x - 50\) \(\equiv (x - 2)(ax^2 + bx + c)\) \(\equiv ax^3 - 2ax^2 + bx^2 - 2bx + cx - 2c\) and any two of \(a = 1\), \(b = 10\), \(c = 25\)	M1
\(x^2 + 10x + 25\)	A1
\((x - 2)(x + 5)^2\)	A1	oe

This work might appear in 17a … you can mark it having seen it in 17a unless there is a contradiction with any work in 17b. Also, mark from what you might see in 17a if there is no work in 17b
Ignore further work which gives answers of 2, \(-5\) and \(-5\) (from solving \(\mathrm{f}(x) = 0\))