Moderate -0.5 This is a straightforward application of the factor and remainder theorems requiring two simultaneous equations. Students substitute x = -1/3 (from the factor) and x = 2 (from the remainder theorem) to find a and b. While it involves some algebraic manipulation with fractions, it's a standard textbook exercise with no novel insight required, making it slightly easier than average.
3 The polynomial \(a x ^ { 3 } + b x ^ { 2 } + x + 3\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). It is given that \(( 3 x + 1 )\) is a factor of \(\mathrm { p } ( x )\), and that when \(\mathrm { p } ( x )\) is divided by \(( x - 2 )\) the remainder is 21 . Find the values of \(a\) and \(b\).
Substitute \(x = -\frac{1}{3}\), equate result to zero or divide by \(3x + 1\) and equate the remainder to zero and obtain a correct equation, e.g. \(-\frac{1}{27}a + \frac{1}{9}b - \frac{1}{3} + 3 = 0\)
B1
Substitute \(x = 2\) and equate result to 21 or divide by \(x - 2\) and equate constant remainder to 21. Obtain a correct equation, e.g. \(8a + 4b + 5 = 21\)
M1
Solve for \(a\) or for \(b\)
M1
Obtain \(a = 12\) and \(b = -20\)
A1
[5]
Substitute $x = -\frac{1}{3}$, equate result to zero or divide by $3x + 1$ and equate the remainder to zero and obtain a correct equation, e.g. $-\frac{1}{27}a + \frac{1}{9}b - \frac{1}{3} + 3 = 0$ | B1 |
Substitute $x = 2$ and equate result to 21 or divide by $x - 2$ and equate constant remainder to 21. Obtain a correct equation, e.g. $8a + 4b + 5 = 21$ | M1 |
Solve for $a$ or for $b$ | M1 |
Obtain $a = 12$ and $b = -20$ | A1 | [5]
3 The polynomial $a x ^ { 3 } + b x ^ { 2 } + x + 3$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( 3 x + 1 )$ is a factor of $\mathrm { p } ( x )$, and that when $\mathrm { p } ( x )$ is divided by $( x - 2 )$ the remainder is 21 . Find the values of $a$ and $b$.
\hfill \mbox{\textit{CAIE P3 2014 Q3 [5]}}