Does the order I multiply the characteristic equation's factors in the homogeneous solution matter? I've been doing a recurrence relation exercise in my book. Doing some steps and comparing them to the ones taken by the book.

$$T(0) = 1$$
$$T(1) = 2$$
$$T(k) - 7T(k-1)+10T(k-2)=6+8k$$


Characteristic equation:
$$a^2-7a+10$$
$$(a - 5)(a-2)$$
Homogeneous solution:
$$b_0\cdot(5)^n+b_1\cdot(2)^n$$
Wait. My book did something different. Its homogeneous solution is
$$b_0\cdot(2)^n+b_1\cdot(5)^n$$
As you can observe, the book multiplied the constants by the factors in a different order as I did.
My question, then: does it matter? If yes, what determines the order?

The order doesn't matter.
A general way of solving recurrences like this is as follows. Define the generating function $G(z) = \sum_{n \ge 0} T(n) z^n$, rewrite the recurrence so there aren't subtractions in indices:
$$
T(n + 2) - 7 T(n + 1) + 10 T(n) = 8 n + 22
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} T(n + 1) z^n &= \frac{G(z) - T(0)}{z} \\
\sum_{n \ge 0} T(n + 2) z^n &= \frac{G(z) - T(0) - T(1) z}{z^2} \\
\sum_{n \ge 0} z^n          &= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n        &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
                            &= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{G(z) - z - 2 z}{z^2} - 7 \frac{G(z) - 1}{z} + 10 G(z)
   = 8 \frac{z}{(1 - z)^2} + 22 \frac{1}{1 - z}
$$
Solve fo $G(z)$, express the result as partial fractions (this is the part where the factorization of the characteristic equation comes up):
\begin{align}
G(z) &= \frac{1 - 7 z + 33 z^2 - 19 z^3}{1 - 9 z + 25 z^2 - 27 z^3 + 10 z^4} \\
     &= \frac{1 - 7 z + 33 z^2 - 19 z^3}{(1 - 5 z) (1 - 2 z) (1 - z)^2} \\
     &= \frac{2}{1 - 5 z} - \frac{9}{1 - 2 z} + \frac{6}{1 - z} + \frac{2}{(1 - z)^2}
\end{align}
Now you can read off the coefficients, if you remember the generalized binomial theorem:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
             = \sum_{k \ge 0} \binom{m + k - 1}{m - 1} u^k
$$
In this case:
\begin{align}
T(n) &= 2 \cdot 5^n - 9 \cdot 2^n + 6 + 2 (n + 1) \\
     &= 2 \cdot 5^n - 9 \cdot 2^n + 2 n + 8
\end{align}
