Generating Function - Methods for obtaining characteristic roots for determining explicit solution

The Question is:

Answer for a) $\begin{align*} f(x)&= & a_0 & + & a_1x & + & a_2x^2 & + & a_3x^3 & + &\cdots \\ -xf(x)&= & & - & a_0x & - & a_1x^2 & -& a_2x^3 & - &\cdots \\ -5x^2f(x)&= & & & & - & 5a_0x^2 & - & 5a_1x^3 & - &\cdots \\ -3x^3f(x)&= & & & & & & - & 3a_0x^3 & - &\cdots \\ \hline \\ \end{align*} \\ f(x)[1-x-5x^2-3x^3]=a_0+x(a_1-a_0)+x^2(a_2-a_1-a_0)+0+\cdots \\ f(x)=\frac{a_0+x(a_1-a_0)+x^2(a_2-a_1-5a_0)}{1-x-5x^2-3x^3} \\$ Answer for b). The technique used is based on this video by Mayur Gohil $\text{ let }a_n=a^n, \text{ then }a^n=a^{n-1}+5a^{n-2}+3a^{n-3} \Rightarrow \frac{a^n}{a^{n-3}}=\frac{a^{n-1}}{a^{n-3}}+\frac{5a^{n-2}}{a^{n-3}}+\frac{3a^{n-3}}{a^{n-3}} \Rightarrow\\ \text{ characteristic polynomial is } a^3-a^2-5a-3=0 \text{ and the characteristic roots are } a=-1 \text { with multiplicity 2 and } a=3 \\ $ $\text{ when }n=0,a_0=1 \Rightarrow 1=A_1(3)^0+(A_2+A_3(0))(-1)^0 \Rightarrow 1=A_1+A_2 \\ \text{ when }n=1,a_1=1 \Rightarrow 1=A_1(3)^1+(A_2+A_3(1))(-1)^1 \Rightarrow 1=3A_1-A_2-A_3 \\ \text{ when }n=2,a_2=6 \Rightarrow 1=A_1(3)^2+(A_2+A_3(2))(-1)^2 \Rightarrow 6=9A_1+A_2+2A_3 \\ $ $ \text{ create an augmented matrix for the system to solve for } A_1,A_2,A_3 \text { using Gaussian elimination } \\ \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & 0 & 1 \\ 3 & -1 & -1 & 1 \\ 9 & 1 & 2 & 6 \end{array}\right) \longrightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & 0 & 9/16 \\ 0 & 1 & 0 & 7/16 \\ 0 & 0 & 1 & 1/4 \end{array}\right) \\ $ $\text{ Explicit solution for } A_1=\frac{9}{16},A_2=\frac{7}{16},A_3=\frac{1}{4} \Rightarrow a_n=\frac{9}{16}(3^n)+\bigl(\frac{7}{16}+\frac{1}{4}(n)\bigr)(-1)^n \Rightarrow \frac{9}{16}3^n+\frac{7}{16}(-1)^n +\frac{1}{4}n(-1)^n \\ \\ $ Answer given in notebook: