Use synthetic division to divide x^3 - 4x^2 + x + 6 by x + 1, then factor the cubic completely.

Step-by-step solution

1. 1

   Convert $x+1$ into the form $x-c$. The corner number in synthetic division is always the zero of the divisor, so rewrite the divisor before you start:

   $x+1=x-(-1)\;\Longrightarrow\; c=-1$

   Using $+1$ here is the single most common mistake with this kind of divisor. The remainder theorem gives the safety net: whatever the table produces must equal $f(-1)=-1-4-1+6=0$.

2. 2

   List the coefficients. All four powers appear in $x^3-4x^2+x+6$, so the top row is

   $1,\;-4,\;1,\;6$

   Keep the signs attached to the coefficients rather than to the operations; sign bookkeeping is what the table automates for you.

3. 3

   Multiply by $-1$ and add, column by column.

   $\begin{array}{r|rrrr}-1 & 1 & -4 & 1 & 6\\ & & -1 & 5 & -6\\ \hline & 1 & -5 & 6 & 0\end{array}$

   Step by step: $1\cdot(-1)=-1$ and $-4+(-1)=-5$; then $-5\cdot(-1)=5$ and $1+5=6$; then $6\cdot(-1)=-6$ and $6+(-6)=0$. Multiplying by a negative $c$ makes the middle row alternate in sign, which is a useful visual check.

4. 4

   Read the quotient and remainder. Dropping one degree, the bottom row $1,-5,6,0$ means

   $\text{quotient}=x^2-5x+6,\qquad \text{remainder}=0$

   so $x^3-4x^2+x+6=(x+1)(x^2-5x+6)$.

5. 5

   Factor the quadratic quotient to finish. The quotient $x^2-5x+6$ needs two numbers with product $6$ and sum $-5$, namely $-2$ and $-3$:

   $x^2-5x+6=(x-2)(x-3)$

   Hence the complete factorization and the roots are

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

   Check the constant term: $(1)(-2)(-3)=6\;\checkmark$, matching the $+6$ in the original cubic.