Remainder Theorem
Using remainder theorem show that $\left(2x + 1\right)$ is a factor of the polynomial $p \left(x\right) = 4x^3 + 4x^2 - x - 1$. Hence factorize the polynomial.
Given: $p \left(x\right) = 4x^3 + 4 x^2 - x -1$
Now, $\;$ $2x + 1 =0$ $\implies$ $x = - \dfrac{1}{2}$
Remainder $=$ Value of $p \left(x\right)$ at $x = - \dfrac{1}{2}$
$\begin{aligned}
p \left(- \dfrac{1}{2}\right) & = 4 \left(- \dfrac{1}{2}\right)^3 + 4 \left(- \dfrac{1}{2}\right)^2 - \left(- \dfrac{1}{2}\right) - 1 \\\\
& = - \dfrac{1}{2} + 1 + \dfrac{1}{2} - 1 \\\\
& = 0
\end{aligned}$
$\implies$ $\left(2x + 1\right)$ is a factor of $p \left(x\right)$
Now,
$\begin{array}{lllll}
2x + 1 & ) & 4x^3 + 4x^2 - x - 1 & ( & 2x^2 + x - 1 \\
& & 4x^3 + 2x^2 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ 2x^2 - x & & \\
& & \ \ \ \ \ \ \ \ 2x^2 + x & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
$\begin{aligned}
\therefore \; 4 x^3 + 4 x^2 - x - 1 & = \left(2x + 1\right) \left(2x^2 + x -1\right) \\\\
& = \left(2x + 1\right) \left(2x^2 + 2 x - x - 1\right) \\\\
& = \left(2x + 1\right) \left[2 x \left(x + 1\right) - 1 \left(x + 1\right)\right] \\\\
& = \left(2x + 1\right) \left(2x - 1\right) \left(x + 1\right)
\end{aligned}$