## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Commence division and reach partial quotient $x^2 + kx$ | M1 | |
| Obtain correct quotient $x^2 + 2x - 1$ | A1 | |
| Set their linear remainder equal to $2x + 3$ and solve for $a$ or $b$ | M1 | Remainder $= (a+3)x + (b-1)$ |
| Obtain answer $a = -1$ | A1 | |
| Obtain answer $b = 4$ | A1 | |

**Alternative method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| State $x^4 + 3x^3 + ax + b = (x^2 + x - 1)(x^2 + Ax + B) + 2x + 3$ and solve two equations in $A$ and $B$ | M1 | e.g. $3 = 1 + A$ and $0 = -1 + A + B$ |
| Obtain $A = 2$, $B = -1$ | A1 | |
| Form and solve equations for $a$ or $b$ | M1 | e.g. $a = B - A + 2$, $b = -B + 3$ |
| Obtain answer $a = -1$ | A1 | |
| Obtain answer $b = 4$ | A1 | |

**Alternative method (remainder theorem):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use remainder theorem with $x = \dfrac{-1 \pm \sqrt{5}}{2}$ | M1 | Allow correct use of either root in exact or decimal form |
| Obtain $-\dfrac{a}{2} \pm \dfrac{a\sqrt{5}}{2} + b = \dfrac{9}{2} \mp \dfrac{\sqrt{5}}{2}$ | A1 | Expand brackets and obtain exact equation for either root. Accept exact equivalent. |
| Solve simultaneous equations for $a$ or $b$ | M1 | |
| Obtain answer $a = -1$ from exact working | A1 | |
| Obtain answer $b = 4$ from exact working | A1 | |

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