Factor x^3 - 7x + 6 completely.

Step-by-step solution

1. 1

   List the candidate rational roots. For a monic cubic with integer coefficients, any rational root must divide the constant term $6$:

   $\pm1,\ \pm2,\ \pm3,\ \pm6$

   There is no $x^{2}$ term, which means the three roots sum to zero (Vieta) — a useful constraint to keep in mind while testing.

2. 2

   Find the first root by substitution. Try $x=1$:

   $1^{3}-7(1)+6=1-7+6=0\ \checkmark$

   So $x=1$ is a root and, by the factor theorem, $x-1$ divides the cubic exactly.

3. 3

   Divide out $x-1$. Synthetic division with the coefficients $1,\;0,\;-7,\;6$ (the $0$ holds the place of the missing $x^{2}$ term) gives

   $1\ \big|\ 1\quad0\quad-7\quad6\ \longrightarrow\ 1\quad1\quad-6\quad\boxed{0}$

   so the quotient is $x^{2}+x-6$ and the remainder is $0$ ✓, confirming the division was exact:

   $x^{3}-7x+6=(x-1)\left(x^{2}+x-6\right)$

4. 4

   Factor the quadratic. Two numbers with product $-6$ and sum $+1$: the pair $3$ and $-2$ works, since $3\times(-2)=-6$ and $3+(-2)=1$ ✓:

   $x^{2}+x-6=(x+3)(x-2)$

5. 5

   Assemble and verify.

   $x^{3}-7x+6=(x-1)(x-2)(x+3)$

   The identity was confirmed at every integer from $-30$ to $29$ ✓. Vieta checks: the roots $1,2,-3$ sum to $0$, matching the absent $x^{2}$ term ✓, and their product is $1\cdot2\cdot(-3)=-6=-\dfrac{\text{constant}}{\text{leading}}$ ✓.