CAIE A-Level Further Math AS 1.1 Roots of Polynomial Equations Questions

$b=-(\alpha+\beta+\gamma+\delta)=-3$$5=(-3)^{2}-2(\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta)$
Uses formula for sum of squares.

c=2
$6=\frac{\alpha \beta \gamma+\beta \gamma \delta+\gamma \delta \alpha+\delta \alpha \beta}{\alpha \beta \gamma \delta}=\frac{-d}{-2}$
Uses
$$\alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1}=\frac{\alpha \beta \gamma+\beta \gamma \delta+\gamma \delta \alpha+\delta \alpha \beta}{\alpha \beta \gamma \delta}.$$

d=12
Equation is $x^{4}-3 x^{3}+2 x^{2}+12 x-2=0$.

6

$\alpha^{4}+\beta^{4}+\gamma^{4}+\delta^{4}=3(-27)-2(5)-12(3)+2(4)$
Uses their quartic equation derived in (a).

-119
2

$$\begin{aligned} y=3 x+1 \Rightarrow x=\frac{1}{3}(y-1) \quad \Rightarrow 27\left(\frac{y-1}{3}\right)^{3}+18\left(\frac{y-1}{3}\right)^{2}+6\left(\frac{y-1}{3}\right)-1=0 \end{aligned}$$
Substitutes.

$$\begin{aligned} \Rightarrow(y-1)^{3}+2(y-1)^{2}+2(y-1)-1=0 \quad \Rightarrow y^{3}-3 y^{2}+3 y-1+2 y^{2}-4 y+2+2 y-2-1=0 \end{aligned}$$
Expands.

$y^{3}-y^{2}+y-2=0$
AG.

3

$S_{2}=1^{2}-2(1)=-1$
Uses formula for sum of squares, AG.

$S_{3}=(3 \alpha+1)^{3}+(3 \beta+1)^{3}+(3 \gamma+1)^{3}=-1-(1)+6$
Uses $y^{3}=y^{2}-y+2$ or expands and
uses original equation.

4
4

$S_{-1}=\frac{(3 \alpha+1)(3 \beta+1)+(3 \beta+1)(3 \gamma+1)+(3 \gamma+1)(3 \alpha+1)}{(3 \alpha+1)(3 \beta+1)(3 \gamma+1)}=\frac{1}{2}$$2 S_{-2}=S_{1}-3+S_{-1}=1-3+\frac{1}{2}$
Uses $2 y^{-2}=y-1+y^{-1}$.

$S_{-2}=-\frac{3}{4}$
CAO

3