Find the remainder when $p (x) = x^{3} + 3 x^{2} + 3 x + 1$ , is divided by $x + π$

- π^{3} + 3 π^{2} - 3 π - 1

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- π^{3} + 3 π^{2} - 3 π + 1

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- π^{3} + 3 π^{2} + 3 π + 1

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π^{3} + 3 π^{2} - 3 π + 1

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Solution

The correct option is B $- π^{3} + 3 π^{2} - 3 π + 1$
Here, $p (x) = x^{3} + 3 x^{2} + 3 x + 1$ , and the zero of $x + π$ is $- π$
So, $p (- π) = (- π)^{3} + 3 \times (- π)^{2} + 3 (- π) + 1$
$= - π^{3} + 3 π^{2} - 3 π + 1$
So, by the Remainder Theorem, $- π^{3} + 3 π^{2} - 3 π + 1$ is the remainder when $x^{3} + 3 x^{2} + 3 x + 1$ is divided by $x + π$ .

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