The problem asks which of the following binomials is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$. The possible binomials are $x-1$, $2x+3$, $x+7$, and $x+1$.

1. Problem Description

The problem asks which of the following binomials is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$. The possible binomials are $x-1$, $2x+3$, $x+7$, and $x+1$.

2. Solution Steps

We can test each binomial to see if it is a factor using the Factor Theorem. The Factor Theorem states that a polynomial $f(x)$ has a factor $(x - a)$ if and only if $f(a) = 0$.

* Test $x-1$:

If $x-1$ is a factor, then $x=1$ is a root. We evaluate the polynomial at $x=1$:

$f(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0$.

Since $f(1) = 0$, $x-1$ is a factor.

* Test $2x+3$:

If $2x+3$ is a factor, then $2x = -3$, so $x = -\frac{3}{2}$ is a root.

$f(-\frac{3}{2}) = (-\frac{3}{2})^3 - 6(-\frac{3}{2})^2 + 11(-\frac{3}{2}) - 6 = -\frac{27}{8} - 6(\frac{9}{4}) - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{54}{4} - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{108}{8} - \frac{132}{8} - \frac{48}{8} = -\frac{315}{8} \neq 0$.

Since $f(-\frac{3}{2}) \neq 0$, $2x+3$ is not a factor.

* Test $x+7$:

If $x+7$ is a factor, then $x = -7$ is a root.

$f(-7) = (-7)^3 - 6(-7)^2 + 11(-7) - 6 = -343 - 6(49) - 77 - 6 = -343 - 294 - 77 - 6 = -720 \neq 0$.

Since $f(-7) \neq 0$, $x+7$ is not a factor.

* Test $x+1$:

If $x+1$ is a factor, then $x=-1$ is a root.

$f(-1) = (-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24 \neq 0$.

Since $f(-1) \neq 0$, $x+1$ is not a factor.

Therefore, $x-1$ is the only binomial in the list that is a factor of the given polynomial.

3. Final Answer

$x-1$