C3 June 2013 Q1
1. Given that\[\frac{3x^4-2x^3-5x^2-4}{x^2-4}\equiv ax^2+bx+c+\frac{dx+e}{x^2-4},\qquad x\neq\pm 2\]find the values of the constants \(a\), \(b\), \(c\), \(d\) and \(e\). (4)
By Division
| Scheme | Marks |
|---|---|
| \[\begin{array}{r}3x^2-2x+7\phantom{-(0x)-4}\\x^2(+0x)-4\,\big)\overline{\,3x^4-2x^3-5x^2+(0x)-4}\\\underline{3x^4+0x^3-12x^2}\phantom{+(0x)-4}\\-2x^3+7x^2+0x\phantom{-4}\\\underline{-2x^3+0x^2+8x}\phantom{-4}\\7x^2-8x-4\\\underline{7x^2+0x-28}\\-8x+24\end{array}\] | |
| \(a=3\) | B1 |
| Long division as far as\[\begin{array}{r}3x^2-2x\ldots\ldots\phantom{-5x^2+(0x)-4}\\x^2(+0x)-4\,\big)\overline{\,3x^4-2x^3-5x^2+(0x)-4}\\\underline{3x^4+0x^3-12x^2}\phantom{+(0x)-4}\\-2x^3+\ldots\ldots\ldots\phantom{(0x)-4}\\\underline{-2x^3+\ldots\ldots\ldots}\phantom{(0x)-4}\end{array}\] | M1 |
| Two of \(b=-2\quad c=7\quad d=-8\quad e=24\) | A1 |
| All four of \(b=-2\quad c=7\quad d=-8\quad e=24\) | A1 |
| (4 marks) |
Notes
B1 Stating \(a=3\). This can also be scored by the coefficient of \(x^2\) in \(3x^2-2x+7\)
M1 Using long division by \(x^2-4\) and getting as far as the ‘\(x\)’ term. The coefficients need not be correct.
Award if you see the whole number part as \(\ldots x^2+\ldots x\) following some working. You may also see this in a table/ grid.
Long division by \((x+2)\) will not score anything until \((x-2)\) has been divided into the new quotient. It is very unlikely to score full marks and the mark scheme can be applied.
A1 Achieving two of \(b=-2\ \ c=7\ \ d=-8\ \ e=24\).
The answers may be embedded within the division sum and can be implied.
A1 Achieving all of \(b=-2\ \ c=7\ \ d=-8\) and \(e=24\)
Accept a correct long division for 3 out of the 4 marks scoring B1M1A1A0
Need to see a=…, b=…, or the values embedded in the rhs for all 4 marks
Alt 1 By Multiplication
| Scheme | Marks |
|---|---|
| \(*\ \ 3x^4-2x^3-5x^2-4\equiv(ax^2+bx+c)(x^2-4)+dx+e\) | |
| Compares the \(x^4\) terms \(a=3\) | B1 |
| Compares coefficients to obtain a numerical value of one further constant \(-2=b,\quad -5=-4a+c\Rightarrow c=..,\) | M1 |
| Two of \(b=-2\quad c=7\quad d=-8\quad e=24\) | A1 |
| All four of \(b=-2\quad c=7\quad d=-8\quad e=24\) | A1 |
| (4 marks) |
B1 Stating \(a=3\). This can also be scored for writing \(3x^4=ax^4\)
M1 Multiply out expression given to get *. Condone slips only on signs of either expression.
Then compare the coefficient of any term (other than \(x^4\)) to obtain a numerical value of one further constant. In reality this means a valid attempt at either \(b\) or \(c\)
The method may be implied by a correct additional constant to \(a\).
A1 Achieving two of \(b=-2\ \ c=7\ \ d=-8\ \ e=24\)
A1 Achieving all of \(b=-2\ \ c=7\ \ d=-8\) and \(e=24\)