Use synthetic division to divide x³ − 2x + 3 by x − 2.

Step-by-step solution

1. 1

   List the coefficients with a placeholder for the missing power. Synthetic division works purely with positional coefficients, so every degree from $3$ down to $0$ needs an entry. Writing the cubic as

   $x^3 + 0x^2 - 2x + 3$

   gives the row $1, \ 0, \ -2, \ 3$. Omitting the $0$ shifts everything and is the classic error here.

2. 2

   Read the synthetic divisor off the linear factor. For a divisor $x - r$ you use $r$ itself, so $x - 2$ means you divide by $2$ (a positive $2$, because $r$ is the root of $x - 2 = 0$). A divisor written $x + 2$ would instead mean $r = -2$.

3. 3

   Run the multiply-and-add passes. Bring down the leading $1$, then repeatedly multiply by $2$ and add to the next coefficient:

   $1 \ \rightarrow\ 0 + 2(1) = 2 \ \rightarrow\ -2 + 2(2) = 2 \ \rightarrow\ 3 + 2(2) = 7$

   The completed tableau is

   $\begin{array}{r|rrrr} 2 & 1 & 0 & -2 & 3 \\ & & 2 & 4 & 4 \\ \hline & 1 & 2 & 2 & 7 \end{array}$

4. 4

   Interpret the bottom row. The last entry is the remainder; the earlier entries are the quotient coefficients, with the degree dropped by one. From $1, 2, 2$ and remainder $7$:

   $\frac{x^3 - 2x + 3}{x - 2} = x^2 + 2x + 2 + \frac{7}{x - 2}$

5. 5

   Verify with the remainder theorem and by multiplying back. The remainder must equal $f(2)$:

   $f(2) = 8 - 4 + 3 = 7 \ \checkmark$

   And expanding the quotient confirms the whole identity:

   $(x-2)(x^2 + 2x + 2) + 7 = x^3 - 2x - 4 + 7 = x^3 - 2x + 3 \ \checkmark$

   Since the remainder is not $0$, $x - 2$ is not a factor of $x^3 - 2x + 3$.