Moderate -0.8 This is a straightforward application of the factor and remainder theorems requiring two substitutions to form simultaneous equations. The arithmetic is routine and the method is standard textbook material with no problem-solving insight needed, making it easier than average.
3 The polynomial \(2 x ^ { 3 } + a x ^ { 2 } - 11 x + b\) is denoted by \(\mathrm { p } ( x )\). It is given that \(\mathrm { p } ( x )\) is divisible by \(( 2 x - 1 )\) and that when \(\mathrm { p } ( x )\) is divided by \(( x + 1 )\) the remainder is 12 .
Find the values of \(a\) and \(b\).
Substitute \(x = \frac{1}{2}\) and equate the result to zero
M1
Or divide as far as \((2x-1)(x^2 + px + q)\) and equate constant remainder to zero
Obtain a correct equation with powers evaluated
A1
e.g. \(\frac{a}{4} + \frac{a}{4} - \frac{11}{2} + b = 0\) or \(a + 4b = 21\)
Substitute \(x = -1\) and equate result to 12
M1
Or divide as far as \((x+1)(2x^2 + rx + s)\) and equate constant remainder to 12
Obtain a correct equation with powers evaluated
A1
e.g. \(-2 + a + 11 + b = 12\) or \(a + b = 3\)
Obtain \(a = -3\), \(b = 6\)
A1
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = \frac{1}{2}$ and equate the result to zero | M1 | Or divide as far as $(2x-1)(x^2 + px + q)$ and equate constant remainder to zero |
| Obtain a correct equation with powers evaluated | A1 | e.g. $\frac{a}{4} + \frac{a}{4} - \frac{11}{2} + b = 0$ or $a + 4b = 21$ |
| Substitute $x = -1$ and equate result to 12 | M1 | Or divide as far as $(x+1)(2x^2 + rx + s)$ and equate constant remainder to 12 |
| Obtain a correct equation with powers evaluated | A1 | e.g. $-2 + a + 11 + b = 12$ or $a + b = 3$ |
| Obtain $a = -3$, $b = 6$ | A1 | |
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3 The polynomial $2 x ^ { 3 } + a x ^ { 2 } - 11 x + b$ is denoted by $\mathrm { p } ( x )$. It is given that $\mathrm { p } ( x )$ is divisible by $( 2 x - 1 )$ and that when $\mathrm { p } ( x )$ is divided by $( x + 1 )$ the remainder is 12 .
Find the values of $a$ and $b$.\\
\hfill \mbox{\textit{CAIE P3 2023 Q3 [5]}}