Use synthetic division to divide x^3 - 4x + 5 by x - 2, and state the quotient and remainder.

Step-by-step solution

1. 1

   Fill in the missing power before you start. The dividend has no $x^2$ term, so write it as

   $x^3+0\cdot x^2-4x+5\;\Longrightarrow\; 1,\;0,\;-4,\;5$

   A cubic always needs four entries in the coefficient row. Omitting the $0$ would make the table compute a division by the wrong polynomial.

2. 2

   Predict the remainder with the remainder theorem. Because the divisor is $x-2$ you have $c=2$, and the remainder of the division must equal the value of the polynomial there:

   $f(2)=2^3-4(2)+5=8-8+5=5$

   So a nonzero remainder of $5$ is expected, and $x-2$ will not be a factor.

3. 3

   Run the table with $c=2$.

   $\begin{array}{r|rrrr}2 & 1 & 0 & -4 & 5\\ & & 2 & 4 & 0\\ \hline & 1 & 2 & 0 & 5\end{array}$

   Bring down $1$; $1\cdot 2=2$ and $0+2=2$; $2\cdot 2=4$ and $-4+4=0$; $0\cdot 2=0$ and $5+0=5$. The predicted remainder $5$ appears in the last column.

4. 4

   Translate the bottom row into a quotient. The entries $1,\,2,\,0$ are the coefficients of a quadratic:

   $\text{quotient}=x^2+2x+0=x^2+2x=x(x+2),\qquad \text{remainder}=5$

   The trailing $0$ means the quotient has no constant term, which is why it factors so cleanly.

5. 5

   Write and check the division identity.

   $\frac{x^3-4x+5}{x-2}=x^2+2x+\frac{5}{x-2}\quad\Longleftrightarrow\quad x^3-4x+5=(x-2)(x^2+2x)+5$

   Expanding the product: $(x-2)(x^2+2x)=x^3+2x^2-2x^2-4x=x^3-4x$, and adding $5$ restores the dividend $\checkmark$. Since the remainder is not zero, $x=2$ is not a root of $x^3-4x+5$.