The clue is in the number 32 which is 2^5. So a=2 may be a factor. It is, because if we set the expression to zero, a=2 would be a solution. This means (a-2) is a factor, so we can use algebraic long division to simplify the equation. When we do this we get a^4+2a^3+8a+16, a quartic. This time because 2^4=16, we might suspect that a=2 is another solution if the expression were zero. But that would mean another factor of (a-2), but we have no minus signs in the expression. Try a=-2 and the middle terms add up to -32 and the outer terms to +32, so (a+2) is another factor. We divide the quartic by (a+2). This time we're left with a^3+8. Clearly (a+2) divides into this because a=-2 makes the expression zero. Now all we're left with is a^2-2a+4 which factorises into (a-2)^2. So totally factorised we have (a-2)(a+2)(a+2)(a-2)(a-2) or (a+2)^2(a-2)^3.