In solving this problem, I found a formula for a product \((1+x_{1})(1+x_{2})(1+x_{3})(1+x_{4})...(1+x_{d})\) up until the last root of a polynomial of degree d. I challenge Brilliant members to find this formula, with proof.

The solution is almost complete. The third time you state \(f\left(-1\right)\), this can be simplified after in the next step. In addition, what if there is a coefficient of the \(x^{n}\) term? Consider these and make an edit.

@Siddhartha Srivastava
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Your solution is just one thing away. Instead of using modular arithmetic in your final formula, what can you do with the sign instead?

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## Comments

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TopNewestit would be kind enough if u tell us......

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There is a pattern.....for example...............

( a + 1 ) ( b + 1 ) ( C + 1 ) = 1 + a + b + c + a b + b c + c a + a b c .

Hope you want this!

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Yes, this is the pattern, but it can be simplified even further. Think about the certain terms in your example.

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Well, Tristan, I can't get it......please post the answer.....

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Let the polynomial be \( f(x) \) with degree \( n \) and roots \( r_1, r_2, \dots, r_{n-1}, r_{n} \)

Suppose \( f(x) = a_0x^n + a_1x^{n-1} + \dots + a_{n-1}x + a_n = a_0(x-r_1)(x-r_2)\dots(x-r_{n-1})(x-r_{n}) \)

Then \( f(-1) = a_0(-1)^n + a_1(-1)^{n-1} + \dots + a_{n-1}(-1) + a_n = a_0(-1-r_1)(-1-r_2)\dots(-1-r_{n-1})(-1-r_n) \)

\( f(-1) = a_0(-1)(1+r_1)(-1)(1+r_2)\dots(-1)(1+r_{n-1})(-1)(1+r_n) \)

\( \dfrac{f(-1)}{a_0} = (-1)^n(1+ r_1)(1 + r_2)\dots(1+ r_{n-1})(1+ r_n) \)

\( (1+ r_1)(1 + r_2)\dots(1+ r_{n-1})(1+ r_n) = \left\{ \begin{array}{lr} \frac{-f(-1)}{a_0} & : n \equiv 1 \pmod2 \\ \frac{f(-1)}{a_0} & : n \equiv 0 \pmod2 \end{array} \right. \)

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The solution is almost complete. The third time you state \(f\left(-1\right)\), this can be simplified after in the next step. In addition, what if there is a coefficient of the \(x^{n}\) term? Consider these and make an edit.

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I don't understand what you meant in the first line. I've edited for the case when there is a coefficient of the leading term.

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I'll let some more people try, then I'll post this answer sometime next week(April 14 to 18).

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It seems that Siddhartha Srivastava has come very close to solving this problem. Congratulations!

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