Question

(a) Suppose $\mathit{r}\mathbf{,}\mathit{s}\mathbf{\in }\mathit{F}$are roots of$\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{\in }\mathit{F}\left[x\right]$ (with $\mathit{a}\mathbf{\ne }{\mathbf{0}}_{\mathbf{F}}$). Use the

Factor Theorem to show that $\mathit{r}\mathbf{+}\mathit{s}\mathbf{=}\mathbf{-}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{b}$and $\mathit{r}\mathit{s}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{c}$.

(b) Suppose $\mathit{r}\mathbf{,}\mathit{s}\mathbf{,}\mathit{t}\mathbf{\in }\mathit{F}$are roots of$\mathit{a}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathit{b}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{x}\mathbf{+}\mathit{d}\mathbf{\in }\mathit{F}\left[x\right]$ (with $\mathit{a}\mathbf{\ne }{\mathbf{0}}_{\mathbf{F}}$)·Show that $\mathit{r}\mathbf{+}\mathit{s}\mathbf{+}\mathit{t}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{b}$and role="math" localid="1648655841662" $\mathit{r}\mathit{s}\mathbf{+}\mathit{s}\mathit{t}\mathbf{+}\mathit{r}\mathit{t}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{c}$and $\mathit{r}\mathit{s}\mathit{t}\mathbf{=}\mathbf{-}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{d}$.

Short answer

It is shown that,

1. $r+s=-a^{-1}b$and $rs=a^{-1}c$.
2. $r+s+t=a^{-1}b$and $rs+st+rt=a^{-1}c$and$rst=-a^{-1}d$

Step-by-step solution

a)Step 1: Finding the proof of r+s=-a-1b and rs=a-1c

Here, r and s are the roots of a polynomial $x^2+a^{-1}bx+a^{-1}c$.

Using the Factor Theorem, then,

$\left(x-r\right)$and $\left(x-s\right)$are the factors of$x^2+a^{-1}bx+a^{-1}c$.

Therefore,

$$\begin{gathered} x^2+a^{-1}bx+a^{-1}c=\left(x-r\right)\left(x-s\right) \\ =x^2-\left(r+s\right)x+rs \end{gathered}$$

Thus, $r+s=-a^{-1}bandrs=a^{-1}c$.

b)Step 2: Determine proof of r+s+t=a-1b and rs+st+rt=a-1c and rst=-a-1d

Here, r, s, and t are roots of the polynomial $x^3+a^{-1}b{x}^2+a^{-1}cx+a^{-1}d$.

Using the Factor Theorem, then,

$\left(x-r\right),\left(x-s\right),and\left(x-t\right)$are the factors of $x^3+a^{-1}b{x}^2+a^{-1}cx+a^{-1}d$.

Therefore,

$$\begin{gathered} x^3+a^{-1}b{x}^2+a^{-1}cx+a^{-1}d=\left(x-r\right)\left(x-s\right)\left(x-t\right) \\ =x^3-\left(r+s+t\right)x^2+\left(rs+st+rt\right)x-rst \end{gathered}$$

Thus, $r+s+t=a^{-1}b$and $rs+st+rt=a^{-1}c$and$rst=-a^{-1}d$

Hence, it is proved.