Use synthetic division to divide x^3 + 3x^2 - 6x - 8 by x - 1. Give the quotient and remainder, and say whether x - 1 is a factor.

Step-by-step solution

1. 1

   Set up with $c=1$. From $x-1=x-(1)$ the corner value is $c=1$, and the coefficient row is

   $1,\;3,\;-6,\;-8$

   Before dividing, predict the remainder with the remainder theorem: $f(1)=1+3-6-8=-10$. Knowing the answer in advance turns the table into a check rather than a leap of faith.

2. 2

   Carry out the multiply-and-add sweep. With $c=1$ the multiplications are trivial, which makes this a good case for seeing the mechanism:

   $\begin{array}{r|rrrr}1 & 1 & 3 & -6 & -8\\ & & 1 & 4 & -2\\ \hline & 1 & 4 & -2 & -10\end{array}$

   Bring down $1$; $1\cdot 1=1$ and $3+1=4$; $4\cdot 1=4$ and $-6+4=-2$; $-2\cdot 1=-2$ and $-8+(-2)=-10$.

3. 3

   Split the bottom row into quotient and remainder. The final entry $-10$ is the remainder; the first three are quotient coefficients of degree $2$:

   $\text{quotient}=x^2+4x-2,\qquad \text{remainder}=-10$

   It matches the predicted $f(1)=-10$, so the arithmetic is sound.

4. 4

   Write the division statement. Any polynomial division can be recorded in two equivalent ways — as a mixed expression or as a product-plus-remainder identity:

   $\frac{x^3+3x^2-6x-8}{x-1}=x^2+4x-2-\frac{10}{x-1}$

   $x^3+3x^2-6x-8=(x-1)(x^2+4x-2)-10$

   The second form is the one to verify, because it involves no fractions.

5. 5

   Answer the factor question and confirm. Because the remainder is $-10\ne 0$, the factor theorem says $x-1$ is not a factor and $x=1$ is not a root. Expanding the identity confirms both numbers:

   $(x-1)(x^2+4x-2)=x^3+4x^2-2x-x^2-4x+2=x^3+3x^2-6x+2$

   and $x^3+3x^2-6x+2-10=x^3+3x^2-6x-8\;\checkmark$. If you needed actual roots you would instead try the divisors of $8$; here $f(1)\ne 0$ simply rules $x=1$ out.