Question

If $\mathit{a}\mathbf{\in }\mathit{F}$is a nonzero root of ${\mathit{C}}_{\mathbf{n}}{\mathit{X}}^{\mathbf{n}}\mathbf{+}{\mathit{C}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{X}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{0}}\mathbf{\in }\mathit{F}\left[x\right]$, show

that${\mathit{a}}^{\mathbf{-}\mathbf{1}}$ is a root of ${\mathit{c}}_{\mathbf{0}}{\mathit{x}}^{\mathbf{n}}\mathbf{+}{\mathit{c}}_{\mathbf{1}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{n}}$.

Short answer

It is shown that, a^-1 is a root of ${\sum }_{i=0}^n{c}_i{a}^{n-i}$.

Step-by-step solution

Determine a-1 is a root of c0xn+c1xn-1+...+cn-1x+cn

Consider the given polynomial, $C_n{X}^n+C_{n-1}{X}^{n-1}+....+c_{1}x+c_0\in F\left[x\right]$.

Therefore, it computes as,

localid="1648625699337" $$\begin{gathered} \left(\sum _{i=0}^n{c}_i{a}^{-\left(n-i\right)}\right)a^n=\sum _{i=0}^n{c}_i{a}^{-n+i+n} \\ =\sum _{i=0}^n{c}_i{a}^i \\ =0 \end{gathered}$$

As $a^n\ne 0$, it concludes that$\sum _{i=0}^n{c}_i{a}^{-\left(n-i\right)}=0$. a is a root of$\sum _{i=0}^n{c}_i{a}^i$ .

Therefore, a^-1 is a root of ${\sum }_{i=0}^n{c}_i{a}^{n-i}$.