Using the binomial theorem,
[tex]\displaystyle (2+x) (3-ax)^4 = 2(3-ax)^4 + x(3-ax)^4 \\\\ (2+x)(3-ax)^4 = 2\sum_{n=0}^4\binom4n 3^{4-n}(-ax)^n + x\sum_{n=0}^4\binom4n 3^{4-n}(-ax)^n \\\\ (2+x)(3-ax)^4 = 2\cdot3^4\sum_{n=0}^4\binom4n 3^{-n}(-a)^nx^n + 3^4\sum_{n=0}^4\binom4n 3^{-n}(-a)^nx^{n+1} \\\\ (2+x)(3-ax)^4 = 162\sum_{n=0}^4\binom4n \left(-\dfrac a3\right)^nx^n + 81\sum_{n=0}^4\binom4n \left(-\dfrac a3\right)^nx^{n+1}[/tex]
The first sum contributes an x³ term when n = 3, and the second sum when n = 2 :
[tex]162\dbinom43\left(-\dfrac a3\right)^3x^3 + 81\dbinom42\left(-\dfrac a3\right)^2 x^{2+1}[/tex]
If the coefficient of the x³ term is 30, then
[tex]162\dbinom43\left(-\dfrac a3\right)^3 + 81\dbinom42\left(-\dfrac a3\right)^2 = 30[/tex]
Solve for a :
[tex]-162\cdot4\cdot\dfrac{a^3}{27} + 81\cdot6\cdot\dfrac{a^2}9 = 30 \\\\ -24a^3 + 54a^2 = 30 \\\\ -4a^3+9a^2=5[/tex]
By inspection, if [tex]\boxed{a=1}[/tex], -4 + 9 = 5, so that's one solution.
Move everything to one side to get
[tex]4a^3-9a^2+5=0[/tex]
Since a = 1 is a solution, the left side must be divisible by (a - 1), so that
[tex]4a^3-9a^2+5 = (a-1)(ba^2+ca+d)[/tex]
for some coefficients b, c, d.
Polynomial division yields
[tex]\dfrac{4a^3-9a^2+5}{a-1} = 4a^2-5a-5[/tex]
Find the remaining solutions (I complete the square here, but feel free to use whichever methods you like best) :
[tex]4a^2-5a-5 = 0 \\\\ 4a^2 - 5a = 5 \\\\ 4\left(a^2-\dfrac54a\right) = 5 \\\\ 4\left(a^2 - \dfrac54 a + \dfrac{25}{64}\right) = 5 + \dfrac{4\cdot25}{64} \\\\ 4 \left(a + \dfrac58\right)^2 = \dfrac{105}{16} \\\\ 2\left(a+\dfrac58\right) = \pm\sqrt{\dfrac{105}{16}} \\\\ a+\dfrac58 = \pm\dfrac{\sqrt{105}}8 \\\\ \boxed{a = \dfrac{-5\pm\sqrt{105}}8}[/tex]
