From Exam Papers

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,

$\begin{array}{lllll} 2x + 1 & ) & 4x^3 + 4x^2 - x - 1 & ( & 2x^2 + x - 1 \\ & & 4x^3 + 2x^2 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ 2x^2 - x & & \\ & & \ \ \ \ \ \ \ \ 2x^2 + x & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{array}$

$\begin{aligned} \therefore \; 4 x^3 + 4 x^2 - x - 1 & = \left(2x + 1\right) \left(2x^2 + x -1\right) \\\\ & = \left(2x + 1\right) \left(2x^2 + 2 x - x - 1\right) \\\\ & = \left(2x + 1\right) \left[2 x \left(x + 1\right) - 1 \left(x + 1\right)\right] \\\\ & = \left(2x + 1\right) \left(2x - 1\right) \left(x + 1\right) \end{aligned}$