It clearly depends on the choice of connection polynomial.
For reference, a Galois Field is formed by creating a "primitive" element. The actual value of which is essentially unimportant -- it acts like the imaginary number does. It is the root of an equation that has neither 1, nor 0 as a root (an irreducible polynomial). In this case, a rule can be written similar to the rule for i^2 = -1. For your case, one possible case would be a^5 = a^4 + a^2 + 1. again, it is very important not to become overly concerned with the "value" of "a". (I might have the connection polynomial incorrect above, I haven't check it rigorously)
now, all GF32 elements can be written as b4*a^4 + b4*a^4 + ... b0*a^0. again, the value of "a" is unimportant, what is important is the rule that a^5 = a^4 + a^2 + 1.
From here, it is a task of multiplication and reduction. a second GF32 element might have an equation c4*a^4 + ... + c0*a^0.
the product of the two elements is c4*b4*a^8 + (c3*b4 + c4*b3)*a^7 + ... a0*b0, where the terms for each power of "a" is the terms of c/b which add up to that power. (4,4; 4,3 3,4; 3,3 4,2 2,4; ...)
Now reduction is done. if a^5 = a^4 + a^2 + 1, then a^6 = a*a^5. likewise a^7 = a^2 * a^5, etc... using this reduction rule will allow the 8th order polynomial to be reduced back to the 4th order one desired.
other methods exist. the above is the easiest to explain.