Factored polynomial x-intercepts (SAT). Game on.

Set $y = 0$: $-(x + 8)(x - 2) = 0$. The factor $-1$ is nonzero, so the zeros come from $x + 8 = 0 \Rightarrow x = -8$ and $x - 2 = 0 \Rightarrow x = 2$. Of the choices, $-8$ is an x-coordinate of an x-intercept.

Set $y = 0$: $4(x + 12)(x - 3)(x + 2) = 0$. The constant $4 \neq 0$, so the zeros are $x = -12$, $x = 3$, and $x = -2$. Among the choices, only $3$ is an x-coordinate of an x-intercept.

Set $y = 0$: $-5(x - 11)(x + 4) = 0$. The constant $-5 \neq 0$, so the zeros are $x = 11$ and $x = -4$. Of the choices, $-4$ is an x-coordinate of an x-intercept.

Set $f(x) = 0$: $2(x - 6)(x - 1)(x + 10) = 0$. The constant $2 \neq 0$, so the zeros are $x = 6$, $x = 1$, and $x = -10$. Among the choices, only $-10$ is an x-coordinate of an x-intercept.

Set $y = 0$: $(x - 15)(x + 7) = 0$. The zeros are $x = 15$ and $x = -7$. Of the choices, $15$ is an x-coordinate of an x-intercept.

Set $y = 0$: $-2(x + 15)(x - 8)(x + 3) = 0$. The constant $-2 \neq 0$, so the zeros are $x = -15$, $x = 8$, and $x = -3$. Among the choices, only $8$ is an x-coordinate of an x-intercept.

Set $f(x) = 0$: $6(x + 9)(x - 13) = 0$. The constant $6 \neq 0$, so the zeros are $x = -9$ and $x = 13$. Of the choices, $-9$ is an x-coordinate of an x-intercept.

Set $y = 0$: $3(x - 5)^2(x + 8) = 0$. The constant $3 \neq 0$, so the zeros are $x = 5$ (from the repeated factor) and $x = -8$. Among the choices, only $5$ is an x-coordinate of an x-intercept.