The remainder of n^4-2n^3-n^2+2n-26 when divi | vedclass

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Question

(C) Let $f(n) = n^4-2n^3-n^2+2n-26$. Factorizing the expression: $f(n) = n^3(n-2) - n(n-2) - 26$ $f(n) = (n^3-n)(n-2) - 26$ $f(n) = n(n^2-1)(n-2) - 26

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(C) Let $f(n) = n^4-2n^3-n^2+2n-26$. Factorizing the expression: $f(n) = n^3(n-2) - n(n-2) - 26$ $f(n) = (n^3-n)(n-2) - 26$ $f(n) = n(n^2-1)(n-2) - 26$ $f(n) = (n-2)(n-1)n(n+1) - 26$. Since $(n-2)(n-1)n(n+1)$ is the product of four consecutive integers,it is always divisible by $4! = 24$. Let $(n-2)(n-1)n(n+1) = 24k$ for some integer $k$. Then $f(n) = 24k - 26$. To find the remainder when divided by $24$,we write: $f(n) = 24k - 48 + 22$ $f(n) = 24(k-2) + 22$. Thus,the remainder is $22$.

The remainder of $n^4-2n^3-n^2+2n-26$ when divided by $24$ is:

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