18N.1.AHL.TZ0.H_8

pestleMathematicsAAHLPaper 118N· ahl-2-12-factor-and-remainder-theorems-sum-and-product-of-rootssource ↗

Consider the equation ${z^4} + a{z^3} + b{z^2} + cz + d = 0$ , where $a$ , $b$ , $c$ , $d \in \mathbb{R}$ and $z \in \mathbb{C}$ .

Two of the roots of the equation are log₂6 and $i\sqrt 3$ and the sum of all the roots is 3 + log₂3.

Show that 6 $a$ + $d$ + 12 = 0.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$- i\sqrt 3$ is a root (A1)

$3 + {\text{lo}}{{\text{g}}_2}3 - {\text{lo}}{{\text{g}}_2}6\left( { = 3 + {\text{lo}}{{\text{g}}_2}\frac{1}{2} = 3 - 1 = 2} \right)$ is a root (A1)

sum of roots: $- a = 3 + {\text{lo}}{{\text{g}}_2}3 \Rightarrow a = - 3 - {\text{lo}}{{\text{g}}_2}3$ M1

Note: Award M1 for use of $- a$ is equal to the sum of the roots, do not award if minus is missing.

Note: If expanding the factored form of the equation, award M1 for equating $a$ to the coefficient of ${z^3}$ .

product of roots: ${\left( { - 1} \right)^4}d$ $= 2\left( {{\text{lo}}{{\text{g}}_2}6} \right)\left( {i\sqrt 3 } \right)\left( { - i\sqrt 3 } \right)$ M1

$= 6\,{\text{lo}}{{\text{g}}_2}6$ A1

Note: Award M1A0 for $d = - 6\,{\text{lo}}{{\text{g}}_2}6$

$6a + d + 12 = - 18 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}6 + 12$

EITHER

$= - 6 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

$= - 6 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

[7 marks]

Examiners’ report

[N/A]