1. **State the problem:** We need to determine which binomial among $x-6$, $x+2$, $x+4$, and $x-7$ is NOT a factor of the polynomial $$x^3 - 11x^2 + 16x + 84.$$\n\n2. **Recall the Factor Theorem:** A binomial $x - a$ is a factor of a polynomial $f(x)$ if and only if $f(a) = 0$.\n\n3. **Test each binomial:**\n- For $x-6$, test $x=6$: $$f(6) = 6^3 - 11(6^2) + 16(6) + 84 = 216 - 396 + 96 + 84 = 0.$$ So $x-6$ is a factor.\n- For $x+2$, test $x=-2$: $$f(-2) = (-2)^3 - 11(-2)^2 + 16(-2) + 84 = -8 - 44 - 32 + 84 = 0.$$ So $x+2$ is a factor.\n- For $x+4$, test $x=-4$: $$f(-4) = (-4)^3 - 11(-4)^2 + 16(-4) + 84 = -64 - 176 - 64 + 84 = -220 \neq 0.$$ So $x+4$ is NOT a factor.\n- For $x-7$, test $x=7$: $$f(7) = 7^3 - 11(7^2) + 16(7) + 84 = 343 - 539 + 112 + 84 = 0.$$ So $x-7$ is a factor.\n\n4. **Conclusion:** The binomial that is NOT a factor is $x+4$.