## Question 4:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Long division: $x^2 + x - 12 \overline{)x^4 + x^3 - 7x^2 + 8x - 48}$, quotient $x^2 + 5$, remainder $3x + 12$ | M1 A1 | M1: Divides to get quadratic quotient and remainder of form $\alpha x + \beta$ ($\alpha, \beta$ not both zero); A1: Correct quotient and remainder |
| $\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12} \equiv x^2 + 5 + \frac{3(x+4)}{(x+4)(x-3)}$ or $\frac{3x+12}{(x+4)(x-3)}$ | M1 | Writes answer as Their Quotient $+ \frac{\text{Their Remainder}}{(x+4)(x-3)}$ |
| $\equiv x^2 + 5 + \frac{3}{(x-3)}$ or states $A = 5,\ B = 3$ | A1 | |

**(4 marks)**

# Question (Partial fractions / algebraic division):

## Alternative: Dividing by linear factors (method 1 - divide by $(x-3)$ first)

| Working | Mark | Notes |
|---------|------|-------|
| $(x^4+x^3-7x^2+8x-48)\div(x-3): Q_1=x^3+4x^2+5x+23, R_1=21$ | M1A1 | M1: first division requires $Q_1$ cubic and $R_1$ constant; second division gives quadratic $Q_2$ and constant $R_2$. A1: Correct quotients and remainders |
| $(x^3+4x^2+5x+23)\div(x+4): Q_2=x^2+5, R_2=3$ | | |
| $\frac{x^4+x^3-7x^2+8x-48}{(x+4)(x-3)}\equiv x^2+5+\frac{3}{x+4}+\frac{21}{(x-3)(x+4)}$ | M1 | Writes answer as $Q_2+\frac{R_2}{x+4}+\frac{R_1}{(x-3)(x+4)}$ |
| $\equiv x^2+5+\frac{3}{(x-3)}$ or states $A=5, B=3$ | A1 | |

## Alternative: Dividing by $(x+4)$ first then $(x-3)$

| Working | Mark | Notes |
|---------|------|-------|
| $(x^4+x^3-7x^2+8x-48)\div(x+4): Q_1=x^3-3x^2+5x-12, R_1=0$ | M1A1 | M1: first division requires $Q_1$ cubic and $R_1$ constant; second gives quadratic $Q_2$ and constant $R_2$. A1: Correct quotients and remainders |
| $(x^3-3x^2+5x-12)\div(x-3): Q_2=x^2+5, R_2=3$ | | |
| $\frac{x^4+x^3-7x^2+8x-48}{(x+4)(x-3)}\equiv x^2+5+\frac{3}{x-3}(+0)$ | M1 | Writes answer as $Q_2+\frac{R_2}{x-3}+\frac{R_1}{(x-3)(x+4)}$ |
| $\equiv x^2+5+\frac{3}{(x-3)}$ or states $A=5, B=3$ | A1 | |

## Alternative: Comparing coefficients

| Working | Mark | Notes |
|---------|------|-------|
| $x^4+x^3-7x^2+8x-48\equiv(x^2+A)(x^2+x-12)+B(x+4)$ | M1 | Multiplies through by $(x^2+x-12)$ to obtain correct lhs and one of $(x^2+A)(x^2+x-12)$ or $B(x+4)$ on rhs; if $(x^2+A)(x^2+x-12)$ expanded must see both $x^2(x^2+x-12)+A(x^2+x-12)$ |
| 2 correct equations e.g. $x^2\Rightarrow A-12=-7$, $x\Rightarrow A+B=8$, const $\Rightarrow -12A+4B=-48$ | A1 | |
| $A=5, B=3$ | M1A1 | M1: Solves to obtain one of $A$ or $B$. A1: Both values correct |

## Alternative: Substitution

| Working | Mark | Notes |
|---------|------|-------|
| $\frac{x^4+x^3-7x^2+8x-48}{x^2+x-12}\equiv x^2+A+\frac{B}{x-3}$ | M1A1 | M1: Substitutes 2 values for $x$. A1: 2 correct equations. **Multiplying through before substitution must satisfy the condition for multiplying through in the previous alternative.** |
| $x=0\Rightarrow 4=A-\frac{B}{3}$, $x=1\Rightarrow\frac{45}{10}=1+A-\frac{B}{2}$ | | |
| $A=5, B=3$ | M1A1 | M1: Solves to obtain one of $A$ or $B$. A1: Both values correct |

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# Question (b) - Differentiation and tangent:

| Working | Mark | Notes |
|---------|------|-------|
| $g'(x)=2x-\frac{3}{(x-3)^2}$ | M1A1ft | M1: $x^2+A+\frac{B}{x-3}\rightarrow 2x\pm\frac{B}{(x-3)^2}$. A1: $x^2+A+\frac{B}{x-3}\rightarrow 2x-\frac{B}{(x-3)^2}$. Follow through their $B$ or the letter $B$ or a made up $B$ |
| **Special case:** If $g(x)$ written as $x^2+5+\frac{3x+12}{(x-3)}$ and correctly attempts to differentiate as $2x$ + quotient rule on $\frac{3x+12}{(x-3)}$ then M mark available but **not** A1ft. Numerator must be linear. | | |
| $g'(4)=2\times4-\frac{3}{(4-3)^2}(=5)$ | M1 | Substitutes $x=4$ into their derivative |
| Uses $m=g'(4)=(5)$ with $(4, g(4))=(4,24)$ to form equation of tangent | M1 | Correct method of finding equation of tangent. Gradient must be $g'(4)$ and point must be attempt on $(4, g(4))$ |
| $y-24=5(x-4)$ | | |
| $y=5x+4$ | A1 | Cso. Mark may be withheld for incorrect "$A$" earlier or any incorrect work leading to correct gradient |

## Alternative to part (b) - first 3 marks (Quotient rule):

| Working | Mark | Notes |
|---------|------|-------|
| $g'(x)=\frac{(x^2+x-12)(4x^3+3x^2-14x+8)-(x^4+x^3-7x^2+8x-48)(2x+1)}{(x^2+x-12)^2}$ | M1A1 | M1: Correct use of quotient rule — must see evidence of $\frac{vu'-uv'}{v^2}$ or formula quoted and attempted. A1: Correct derivative |
| $g'(4)=\frac{8\times256-192\times9}{8^2}(=5)$ | M1 | Substitutes $x=4$ into their derivative |

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