Let f(x) = n^2 - 1, where n is an integer and n >= 1. Is f(x) always divisible by (x - 1)? Justify.
Accepted Solution
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Answer:Yes.Step-by-step explanation:(Assume x is not 1.)[tex]\frac{x^n-1}{x-1}[/tex] is always divisible for integers greater than 1.Let's use synthetic division:1 | 1x^n + 0x^(n-1) +0x^(n-2) + ....+0x^3+0x^2+0x -1 | 1 1 1 1 1 1 --------------------------------------------------------------------------------- 1 1 1 1 1 1 0We see the remainder is 0 which means that [tex](x-1)[/tex] divides [tex](x^n-1)[/tex].The quotient is [tex]x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^2+x+1[/tex].