I've factorised the equation to $x(x+1)(x-1)(2x-1)(x+3)$ and I notice it has $3$ consecutive numbers, a missing one and then the next consecutive number not sure how to use this to prove the required divisibility.
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For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Henrik supports the community Feb 10 '24 at 09:33
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2You've got to edit this and make clear that you mean that $x$ is an integer. – ancient mathematician Feb 10 '24 at 09:38
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There were no restrictions for x in the question and it was an exam question so I'm not sure either. – Vansh Kumar Feb 10 '24 at 09:39
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Bad question in that case. It makes no sense otherwise. The polynomial is divisible by $5$ if we allow rational coefficients, but not if we insist on integer coefficients. – ancient mathematician Feb 10 '24 at 09:40
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Hint: $x=5k+r$ with $r=0,1,2,3,4$. Deal with these five cases. – ancient mathematician Feb 10 '24 at 09:42
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$!!\bmod 5!:\ 2x-1 \equiv 2(x+2),$ so the product is congruent to twice the product of $5$ consecutive integers, so it is $\equiv 0,$ by the linked dupe. $\ \ $ – Bill Dubuque Feb 10 '24 at 15:08
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as you says, we have $$(x-1)x(x+1)(x+3) (2x-1) ~,$$ so we just need to proof that: when $(x-1)x(x+1)(x+3) \not \equiv 0\pmod 5$, $2x-1 \equiv 0 \pmod 5$.
$(x-1)x(x+1)(x+3) \not \equiv 0\pmod 5$ means that $x+2 \equiv 0 \pmod 5$ so $x \equiv 3 \pmod 5$.then $2x-1 \equiv 6-1 \equiv 0 \pmod 5$.
Q.E.D
int256
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Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Feb 10 '24 at 15:08