When the polynomial $x^{3} + 3 x^{2} + 3 x + 1$ is divided by $x + 1,$ the remainder is

1

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8

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0

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- 6

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Solution

The correct option is C $0$
$\begin{matrix} w e h a v e t o d i v i d e x^{3} + 3 x^{2} + 3 x + 1 b y x + 1 N o w l o n g d i v i s i o n m e t h o d x + 1) x^{3} + 3 x^{2} + 3 x + 1 (x^{2} + 2 x + 1 x^{3} + x^{2} - - - - - - - - - - - - - - 2 x^{2} + 3 x + 1 2 x^{2} + 2 x - - - - - - - - - - - - - - x + 1 x + 1 - - - - - - - - - - - - - - 0 T h e r e m a i n d e r i s 0 H e n c e, t h e o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

Find the remainder when the polynomial $x^{3} + 3 x^{2} + 3 x + 1$ is divided by
$(x - \frac{1}{2})$ .

x^{3} - 3 x^{2} + x + 1

x - 1

The remainder when $x^{3}$ +3 $x^{2}$ +3x+1 is divided by x+1 is

x^{3} + 3 x^{2} + 3 x + 1

x + 1

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