Use synthetic division to divide x^3 - 2x^2 - 4x + 8 by x - 2. Give the quotient and the remainder.

Step-by-step solution

1. 1

   Read off the value of $c$ from the divisor. Synthetic division only works when the divisor is written as $x-c$, and the number you place in the corner is the root of that divisor, not the number you see in the expression. Here $x-2=x-(2)$, so

   $c=2$

   A common slip is to use $-2$ because of the minus sign printed in $x-2$. Test it against the remainder theorem instead: the remainder must equal $f(c)$, and $f(2)=8-8-8+8=0$, which is what the table should produce.

2. 2

   Write the coefficients in descending order, inserting zeros for missing powers. For $x^3-2x^2-4x+8$ every power from $x^3$ down to $x^0$ is present, so nothing has to be padded:

   $1,\;-2,\;-4,\;8$

   This step is where most errors enter. If a power is absent you must still reserve a slot with a $0$, otherwise every later multiply-and-add lands one column too far left.

3. 3

   Run the multiply-and-add cycle. Bring the leading $1$ straight down, then repeatedly multiply the number just written by $c=2$ and add it to the next coefficient:

   $\begin{array}{r|rrrr}2 & 1 & -2 & -4 & 8\\ & & 2 & 0 & -8\\ \hline & 1 & 0 & -4 & 0\end{array}$

   Concretely: $1\cdot 2=2$ and $-2+2=0$; then $0\cdot 2=0$ and $-4+0=-4$; then $-4\cdot 2=-8$ and $8+(-8)=0$.

4. 4

   Interpret the bottom row. The last entry is the remainder and everything to its left lists the quotient coefficients, with the degree dropped by one because you divided by a linear factor:

   $\text{quotient}=1x^2+0x-4=x^2-4,\qquad \text{remainder}=0$

   The $0$ in the middle column is not decoration — it is the coefficient of $x$, which is why the quotient has no linear term.

5. 5

   Check by multiplying back. A remainder of $0$ means the division was exact, so the product of divisor and quotient must rebuild the original cubic:

   $(x-2)(x^2-4)=x^3-4x-2x^2+8=x^3-2x^2-4x+8\;\checkmark$

   Because the remainder vanished you have also factored the polynomial, and $x^2-4$ splits further into $(x-2)(x+2)$, giving $x^3-2x^2-4x+8=(x-2)^2(x+2)$ and roots $x=2$ (double) and $x=-2$.