Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution of x = -1 into p(x) = 0, then rearranging to express m in terms of k. It's a single-step problem with routine algebraic manipulation, making it easier than average for A-level.
1 The polynomial \(\mathrm { p } ( x )\) is defined by
$$\mathrm { p } ( x ) = 4 x ^ { 3 } + ( k + 1 ) x ^ { 2 } - m x + 3 k$$
where \(k\) and \(m\) are constants. Given that \(( x + 1 )\) is a factor of \(\mathrm { p } ( x )\), express \(m\) in terms of \(k\).
Substitute \(-1\) into \(p(x)\) and equate to zero
M1
Allow algebraic long division or the use of an identity with the remainder, in terms of \(m\) and \(k\), equated to zero
Obtain \(-4 + (k+1) + m + 3k = 0\) or equivalent
A1
Obtain \(m = 3 - 4k\)
A1
Total
3
**Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $-1$ into $p(x)$ and equate to zero | M1 | Allow algebraic long division or the use of an identity with the remainder, in terms of $m$ and $k$, equated to zero |
| Obtain $-4 + (k+1) + m + 3k = 0$ or equivalent | A1 | |
| Obtain $m = 3 - 4k$ | A1 | |
| **Total** | **3** | |
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1 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 4 x ^ { 3 } + ( k + 1 ) x ^ { 2 } - m x + 3 k$$
where $k$ and $m$ are constants. Given that $( x + 1 )$ is a factor of $\mathrm { p } ( x )$, express $m$ in terms of $k$.\\
\hfill \mbox{\textit{CAIE P2 2019 Q1 [3]}}