Moderate -0.8 This is a straightforward application of the factor theorem requiring students to substitute x=2 and x=-1/2 into p(x)=0, then solve the resulting simultaneous equations for a and b. It's a standard textbook exercise with clear methodology and no conceptual challenges, making it easier than average.
2 The polynomial \(\mathrm { p } ( x )\) is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + a x ^ { 2 } + 9 x + b$$
where \(a\) and \(b\) are constants. It is given that \(( x - 2 )\) and \(( 2 x + 1 )\) are factors of \(\mathrm { p } ( x )\).
Find the values of \(a\) and \(b\).
Substitute \(x = -\frac{1}{2}\) and equate to zero
M1
Obtain \(4a + b + 66 = 0\) and \(\frac{1}{4}a + b - \frac{21}{4} = 0\) or equivalents
A1
Solve a relevant pair of linear simultaneous equations
DM1
Dependent on at least one M mark
Obtain \(a = -19\), \(b = 10\)
A1
Total
5
**Question 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = 2$ and equate to zero | M1 | |
| Substitute $x = -\frac{1}{2}$ and equate to zero | M1 | |
| Obtain $4a + b + 66 = 0$ and $\frac{1}{4}a + b - \frac{21}{4} = 0$ or equivalents | A1 | |
| Solve a relevant pair of linear simultaneous equations | DM1 | Dependent on at least one M mark |
| Obtain $a = -19$, $b = 10$ | A1 | |
| **Total** | **5** | |
2 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + a x ^ { 2 } + 9 x + b$$
where $a$ and $b$ are constants. It is given that $( x - 2 )$ and $( 2 x + 1 )$ are factors of $\mathrm { p } ( x )$.\\
Find the values of $a$ and $b$.\\
\hfill \mbox{\textit{CAIE P2 2020 Q2 [5]}}