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Find the remainder when$x^3 + 3 x ^2 + 3 x + 1 \ divided \ by 'x'$ by long divison method
Solution:
$x^3 + 3 x ^2 + 3 x + 1 \ divided \ by 'x'$
$ x ) x^ 3 + 3 x^ 2 + 3 x + 1 ( x^2 + 3 x + 3$
$ x ^3$
________________________
$0 + 3 x^ 2 $
$3 x^ 2 $
________________________
$0 + 3 x $
$ 3 x$
________________________
$0 + 1$
So, 1 is the remainder.