Hey there! Great job factoring the polynomial down to x(x+1)(x-1)(2x-1)(x+3) — that's the perfect starting point for this proof. Let's break down how to use those factors to show the product is always divisible by 5, no matter what integer x you pick.

First, let's rephrase one of the factors to make the pattern clearer: x+3 is equivalent to x - 2 when we're working modulo 5 (since 3 = 5 - 2, so subtracting 2 is the same as adding 3 in mod 5 arithmetic). Now our factors look like:
(x-2)(x-1)x(x+1)(2x-1)

Notice the first four terms: (x-2), (x-1), x, (x+1) are four consecutive integers. For any integer x, these four numbers cover four consecutive residues modulo 5. That means either:

• One of these four consecutive numbers is a multiple of 5 (so the whole product is divisible by 5), OR
• The four consecutive numbers are exactly the residues 1,2,3,4 modulo 5 (in some order). In this case, let's check the fifth factor 2x-1:
  • If the four terms are 1,2,3,4, that means x ≡ 3 mod 5 (since x-2=1 → x=3). Plugging x=3 into 2x-1 gives 2*3 -1 = 5, which is a multiple of 5.

Alternatively, we can just test all possible residues of x modulo 5 (since integers repeat every 5 numbers, we only need to check 0,1,2,3,4):

• If x ≡ 0 mod5: The factor x is 0 mod5 → product ≡ 0 mod5
• If x ≡1 mod5: The factor x-1 is 0 mod5 → product ≡ 0 mod5
• If x ≡2 mod5: The factor x+3 = 2+3=5 ≡0 mod5 → product ≡ 0 mod5
• If x ≡3 mod5: The factor 2x-1 = 6-1=5 ≡0 mod5 → product ≡ 0 mod5
• If x ≡4 mod5: The factor x+1=4+1=5≡0 mod5 → product ≡ 0 mod5

In every single case, the product is congruent to 0 modulo 5 — which means the original polynomial is always divisible by 5, no matter what integer x you choose.

备注：内容来源于stack exchange，提问作者Vansh Kumar