Easy -1.2 This is a straightforward application of the factor theorem requiring substitution of x=3 into g(x), setting equal to zero, and solving a linear equation for k. It's a single-step problem with routine algebraic manipulation, making it easier than average A-level questions.
1.
$$g ( x ) = 3 x ^ { 3 } - 20 x ^ { 2 } + ( k + 17 ) x + k$$
where \(k\) is a constant.
Given that \(( x - 3 )\) is a factor of \(\mathrm { g } ( x )\), find the value of \(k\).
Attempts \(g(3)=0\) to set up linear equation in \(k\). Expect to see 3 substituted for \(x\) at least twice. Missing brackets may be recovered.
\(4k - 48 = 0 \Rightarrow k = \ldots\)
M1
Scored for attempting to solve a linear equation in \(k\) having attempted \(g(\pm3)=0\)
\(\{k=\}12\)
A1
Obtains \(\{k=\}12\) only. Do not accept e.g. \(\frac{48}{4}\). Allow slips in working to be recovered. Condone e.g. \(x=12\) provided it comes from a linear equation in \(k\).
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(3) = 3(3)^3 - 20(3)^2 + 3(k+17) + k = 0$ | M1 | Attempts $g(3)=0$ to set up linear equation in $k$. Expect to see 3 substituted for $x$ at least twice. Missing brackets may be recovered. |
| $4k - 48 = 0 \Rightarrow k = \ldots$ | M1 | Scored for attempting to solve a linear equation in $k$ having attempted $g(\pm3)=0$ |
| $\{k=\}12$ | A1 | Obtains $\{k=\}12$ only. Do not accept e.g. $\frac{48}{4}$. Allow slips in working to be recovered. Condone e.g. $x=12$ provided it comes from a linear equation in $k$. |
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1.
$$g ( x ) = 3 x ^ { 3 } - 20 x ^ { 2 } + ( k + 17 ) x + k$$
where $k$ is a constant.\\
Given that $( x - 3 )$ is a factor of $\mathrm { g } ( x )$, find the value of $k$.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q1 [3]}}