For n 2 and n in N,the product of the roots | vedclass

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(C) Let $y = x^2 - 2nx$. The equation becomes $(x-n)(y^2 + (2n^2-5)y + (n^4-5n^2+4)) = 0$. Factoring the quadratic in $y$: $y^2 + (2n^2-5)y + (n^2-1)(

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Let $y = x^2 - 2nx$. The equation becomes $(x-n)(y^2 + (2n^2-5)y + (n^4-5n^2+4)) = 0$. Factoring the quadratic in $y$: $y^2 + (2n^2-5)y + (n^2-1)(n^2-4) = (y + n^2-1)(y + n^2-4) = 0$. Substituting $y$ back: $(x-n)(x^2 - 2nx + n^2 - 1)(x^2 - 2nx + n^2 - 4) = 0$. This simplifies to $(x-n)((x-n)^2 - 1)((x-n)^2 - 4) = 0$. $(x-n)(x-n-1)(x-n+1)(x-n-2)(x-n+2) = 0$. The roots are $x = n, n+1, n-1, n+2, n-2$. The product of the roots $P = n(n+1)(n-1)(n+2)(n-2) = (n-2)(n-1)n(n+1)(n+2)$. This is the product of $5$ consecutive integers,which is always divisible by $5! = 120$.

For $n > 2$ and $n \in N$,the product of the roots of $(x-n)((x^2-2nx)^2 + (2n^2-5)(x^2-2nx) + (n^4-5n^2+4)) = 0$ is divisible by

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