Apply the transformation $ x \rightarrow x + 1$ we have the zero-coefficient is $ 3^2(3.2 + 2^3 + 2^2 + 2^2.3)$ which is divisible by 2 and not divisible by 4.All of the coefficients of $ x,x^2,x^3$ are divisible by 2.Apply the Generalization of Eisenstein's Criterion,then if $ P(x) = x^5 + b_3x^3 + b_2x^2 + b_1x + b_0$ is reducible,it is divisible by a polynomial $ R(x)$ of integer coefficients which $ deg(R) \geq 4$.So $ P(x)$ has a rational root,then it has a integer root because $ P(x)$ is monic,call this root $ x_0$.Set $ x_0 = 2^p.3^q.r$(easy to see that $ r =-1$),check all case of permutations we will have a contradiction. (sorry because of my bad English and explaining)

When you apply the transformation x --->x+1 to $ P(x) = x^5 + b_3x^3 + b_2x^2 + b_1x + b_0$ then $ P(x+1)=(x+1)^5+b_3(x+1)^3+b_2(x+1)^2+b_1(x+1)+b_0$ the coefficient constant is $ 1+b_3+b_2+b_1+b_0=1+54+72+36+108=271$ and not $ 3^2(3.2 + 2^3 + 2^2 + 2^2.3)=270$

It's pretty easy to check that the polynomial cannot have any integral roots (just use Rational Root Theorem and do some case bashing). Hence, we can let $x^5 + b_3x^3 + b_2x^2 + b_1x + b_0 = (x^2 + ax + b)(x^3 + cx^2 + dx + e).$ Right off the bat, we can let $c = -a$ by looking at the $x^4-$coefficient. We now have that $\{b + d - a^2, e+ad - ab, ae+bd, be\} = \{36, 54, 72, 108\}.$ If $3 \nmid b$ or $3 \nmid e$, we have a contradiction from a similar argument as Eisenstein. Else, we have $3 | b, e$. This means that $3| ad, d - a^2$ which implies $3| a, d$. This means that $3| a, b, d, e.$ Since $9| e + ad - ab$, we have $9 | e$ and so $27 | be$. This means that $be = 54$ or $108$. If $be = 54$, then we finish from Eisenstein with $p = 2.$ Hence we have $be = 108.$ If $2 \nmid b$ or $2 \nmid e$, we finish from a similar argument as Eisenstein. Else we have $2| b, e$ and so $e = \pm 18, b = \pm 6.$ From here, the problem is a finite case check (it may first help to get $e + ad - ab = 54$ to cut the search space approximately in third). $\square$