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

Given:

$x^3+ 3x^2 + 3x + 1$ is divided by $x-\frac{1}{2}$

To do:

We have to find the remainder when $x^3+ 3x^2 + 3x + 1$ is divided by $x-\frac{1}{2}$.

Solution:

The remainder theorem states that when a polynomial, $p(x)$ is divided by a linear polynomial, $x - a$ the remainder of that division will be equivalent to $p(a)$.

Let $f(x) = x^3+ 3x^2 + 3x + 1$ and $g(x) = x-\frac{1}{2}$

So, the remainder will be $f(\frac{1}{2})$.

$f(\frac{1}{2}) = (\frac{1}{2})^3+3(\frac{1}{2})^2+3(\frac{1}{2}) + 1$

$= \frac{1}{8} + 3(\frac{1}{4}) +\frac{3}{2}+1$

$=\frac{1+3(2)+3(4)+1(8)}{8}$

$=\frac{1+6+12+8}{8}$

$=\frac{27}{8}$

Therefore, the remainder is $\frac{27}{8}$.