To determine which binomial is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$, we can use the Factor Theorem. This theorem states that if $x - c$ is a factor of a polynomial, then the polynomial will equal zero when $x = c$.

We can test each option by substituting the values into the polynomial.

1. Option a: $x - 1$

   • Substitute $x = 1$: $1^3 - 6(1^2) + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$
   • Factor: Yes

2. Option b: $x + 7$

   • Substitute $x = -7$: $(-7)^3 - 6(-7)^2 + 11(-7) - 6 = -343 - 294 - 77 - 6 \neq 0$
   • Factor: No

3. Option c: $x + 1$

   • Substitute $x = -1$: $(-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 \neq 0$
   • Factor: No

4. Option d: $2x + 3$

   • Substitute $x = -\frac{3}{2}$: $\left(-\frac{3}{2}\right)^3 - 6\left(-\frac{3}{2}\right)^2 + 11\left(-\frac{3}{2}\right) - 6$
   • This computation is lengthy, and evaluating the polynomial doesn't yield zero.

After checking, the correct answer is:

a) $x - 1$

If you would like further details or have any questions, let me know! Here are some related questions you might consider:

1. What are the other factors of the polynomial?
2. How can synthetic division be used to factor the polynomial further?
3. What does the graph of the polynomial look like?
4. How can you confirm the factorization is correct?
5. What is the significance of the roots of a polynomial?

Tip: Always start by testing rational roots that are factors of the constant term!