Find the remainder when ${x^3} + 3{x^2} + 3x + 1$ is divided by $x + \pi $.
A. $\pi $
B. $ - {\pi ^2} + 3{\pi ^3} - 3\pi + 1$
C. $ - {\pi ^3} + 3{\pi ^2} - 3\pi + 1$
D. None of these

Hint: We will use the remainder theorem to find the remainder and write the statement of the remainder theorem to do the same.

Complete step-by-step solution:
Let us first of all mention what the remainder theorem says:-
Remainder theorem: The Remainder Theorem states that if a polynomial f (x) is divided by (x - k) then the remainder r = f (k), where r is the remainder.
If we compare this to the given situation to us with this remainder theorem:-
We have f(x) = ${x^3} + 3{x^2} + 3x + 1$ and k = $ - \pi $.
So, the required remainder should be $f\left( { - \pi } \right)$.
Since, $f(x) = {x^3} + 3{x^2} + 3x + 1$
Therefore, \[f\left( { - \pi } \right) = {\left( { - \pi } \right)^3} + 3{\left( { - \pi } \right)^2} + 3\left( { - \pi } \right) + 1\]
We know that the cube of negative is also negative but the square of negative is always positive.
\[ \Rightarrow f\left( { - \pi } \right) = - {\pi ^3} + 3{\pi ^2} - 3\pi + 1\]

Hence, the required correct answer is (C).

Note: The students must note that they may also try the long division method. But it will create a lot of calculations and hassles which may create calculation errors and everything. So, the remainder theorem works best for this kind of question.
The students must also take care that they might make the mistake of not considering the negative sign while comparing the statement of Remainder Theorem with the given question and that will bring out an incorrect answer.
The students must also know that there is a corollary of Remainder Theorem as well which helps arise the Factor theorem which states that for any polynomial f(x), if f(a) is zero then, (x – a) is a factor of f(x). Because we see that f(a) = 0, which means that remainder is zero and if remainder is 0 when we divide the polynomial by (x – a) that means a must be the root of the polynomial f(x).