Moderate -0.8 This is a straightforward application of the factor theorem with the factor already given. Students divide by (x-3) using polynomial division or inspection, then solve the resulting quadratic—a routine multi-step procedure requiring no problem-solving insight, making it easier than average despite being Further Maths content.
1.
$$f ( x ) = 2 x ^ { 3 } - 8 x ^ { 2 } + 7 x - 3$$
Given that \(x = 3\) is a solution of the equation \(\mathrm { f } ( x ) = 0\), solve \(\mathrm { f } ( x ) = 0\) completely.
(5)
Attempt to expand \((z-(1+2i))(z-(1-2i))\) or any valid method to establish quadratic factor. e.g. \(z = 1 \pm 2i \Rightarrow z-1 = \pm 2i \Rightarrow z^2-2z+1=-4\); or \(z = 1\pm\sqrt{-4} = \frac{2\pm\sqrt{-16}}{2} \Rightarrow b=-2, c=5\); Sum of roots 2, product of roots 5 \(\therefore z^2-2z+5\)
\(f(z) = (z^2-2z+5)(2z+1)\)
M1
Attempt at linear factor with their \(cd\) in \((z^2+az+c)(2z+d) = \pm 5\). Or \((z^2-2z+5)(2z+a) \Rightarrow 5a=5\)
\(z_3 = -\frac{1}{2}\)
A1
Total: 5 marks
# Question 1:
$f(z) = 2z^3 - 3z^2 + 8z + 5$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - 2i$ (is also a root) | B1 | seen |
| $(z-(1+2i))(z-(1-2i)) = z^2 - 2z + 5$ | M1A1 | Attempt to expand $(z-(1+2i))(z-(1-2i))$ or any valid method to establish quadratic factor. e.g. $z = 1 \pm 2i \Rightarrow z-1 = \pm 2i \Rightarrow z^2-2z+1=-4$; or $z = 1\pm\sqrt{-4} = \frac{2\pm\sqrt{-16}}{2} \Rightarrow b=-2, c=5$; Sum of roots 2, product of roots 5 $\therefore z^2-2z+5$ |
| $f(z) = (z^2-2z+5)(2z+1)$ | M1 | Attempt at **linear** factor with their $cd$ in $(z^2+az+c)(2z+d) = \pm 5$. Or $(z^2-2z+5)(2z+a) \Rightarrow 5a=5$ |
| $z_3 = -\frac{1}{2}$ | A1 | |
**Total: 5 marks**
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1.
$$f ( x ) = 2 x ^ { 3 } - 8 x ^ { 2 } + 7 x - 3$$
Given that $x = 3$ is a solution of the equation $\mathrm { f } ( x ) = 0$, solve $\mathrm { f } ( x ) = 0$ completely.\\
(5)\\
\hfill \mbox{\textit{Edexcel FP1 Q1}}