1. Find the remainder when $$x^3 + 3x^2 + 3x + 1$$ is divided by
i)$$x + 1$$
ii)$$x - 1/2$$
iii)$$x$$
iv)$$x + \pi$$
v)$$5 + 2x$$

Let $$p(x) = x^3 + 3x^2 + 3x + 1$$
i)The zero of $$x + 1$$ is x = -1 i.e. (x + 1 = 0)
So, $$p(-1) = {-1}^3 + 3{-1}^2 + 3{-1} + 1$$
i.e. $$p(-1) = -1 + 3 - 3 +1$$
i.e.$$p(-1) = 0$$
Hence, By remainder theorem, required remainder = 0.

ii)The zero of $$x - 1/2$$ is x = 1/2 i.e. (x - 1/2 = 0)
So, $$p(1/2) = {1/2}^3 + 3{1/2}^2 + 3{1/2} + 1$$
i.e. $$p(1/2) = 1/8 + 3/4 + 3/2 +1$$
i.e.$$p(1/2) = 27/8$$
Hence, By remainder theorem, required remainder = 27/8.

iii)The zero of $$x$$ is x = 0 i.e. (x = 0)
So, $$p(0) = {0}^3 + 3{0}^2 + 3{0} + 1$$
i.e. $$p(-1) = 1$$
Hence, By remainder theorem, required remainder = 1.

iv)The zero of $$x + \pi$$ is $$x = -\pi$$ i.e. $$(x + \pi = 0)$$
So, $$p(-\pi) = {-\pi}^3 + 3{-\pi}^2 + 3{-\pi} + 1$$
i.e. $$p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1$$
Hence, By remainder theorem, required remainder = $$-\pi^3 + 3\pi^2 - 3\pi + 1$$.

v)The zero of $$5 + 2x$$ is x = -5/2 i.e. (5 + 2x = 0)
So, $$p(-5/2) = {-5/2}^3 + 3{-5/2}^2 + 3{-5/2} + 1$$
i.e. $$p(-5/2) = -125/8 + 75/4 - 15/2 + 1$$
i.e. $$p(-5/2) = -27/8$$
Hence, By remainder theorem, required remainder = -27/8.