Use synthetic division to divide x^3 - x^2 - 4 by x - 2. Note that the x term is missing.

Step-by-step solution

1. 1

   Spot the gap in the polynomial. Written out with every power present, the dividend is

   $x^3-x^2+0\cdot x-4$

   so the coefficient row must be $1,\;-1,\;0,\;-4$ — four entries for a cubic. Skipping the $0$ and writing only $1,-1,-4$ shifts all later columns and is the classic way this problem goes wrong.

2. 2

   Take $c=2$ from the divisor. Since $x-2=x-(2)$, the corner value is $c=2$. The remainder theorem predicts what the last column must be:

   $f(2)=8-4-4=0$

   so expect an exact division.

3. 3

   Multiply and add across the padded row.

   $\begin{array}{r|rrrr}2 & 1 & -1 & 0 & -4\\ & & 2 & 2 & 4\\ \hline & 1 & 1 & 2 & 0\end{array}$

   In detail: $1\cdot 2=2$ with $-1+2=1$; $1\cdot 2=2$ with $0+2=2$ (this is the column the placeholder made possible); $2\cdot 2=4$ with $-4+4=0$.

4. 4

   Read off the result. The bottom row gives a quadratic quotient and a zero remainder:

   $\text{quotient}=x^2+x+2,\qquad \text{remainder}=0$

   so $x-2$ is a factor and $x=2$ is a root of the cubic.

5. 5

   Check the factorization and the remaining roots. Expanding back,

   $(x-2)(x^2+x+2)=x^3+x^2+2x-2x^2-2x-4=x^3-x^2-4\;\checkmark$

   The quotient $x^2+x+2$ has discriminant $1-8=-7<0$, so it does not factor over the reals. That means $x=2$ is the only real root, and the other two are the complex pair $x=\tfrac{-1\pm i\sqrt7}{2}$.