Find $p(1)$,$p(2)$,and $p(4)$ for the polynomial $p(x) = x^{3} - 7x^{2} + 14x - 8$.

To find the values,we substitute $x = 1$,$x = 2$,and $x = 4$ into the polynomial $p(x) = x^{3} - 7x^{2} + 14x - 8$.
$1$. For $p(1)$:
$p(1) = (1)^{3} - 7(1)^{2} + 14(1) - 8 = 1 - 7 + 14 - 8 = 0$.
$2$. For $p(2)$:
$p(2) = (2)^{3} - 7(2)^{2} + 14(2) - 8 = 8 - 7(4) + 28 - 8 = 8 - 28 + 28 - 8 = 0$.
$3$. For $p(4)$:
$p(4) = (4)^{3} - 7(4)^{2} + 14(4) - 8 = 64 - 7(16) + 56 - 8 = 64 - 112 + 56 - 8 = 0$.
Thus,$p(1) = 0$,$p(2) = 0$,and $p(4) = 0$.