Find $p(1)$,$p(2)$,and $p(4)$ for the following polynomial:
$p(x) = x^{2} - 3x + 2$
$p(x) = x^{2} - 3x + 2$
(N/A) To find the values of the polynomial $p(x) = x^{2} - 3x + 2$ at $x = 1, 2, 4$,we substitute the values into the expression:
$1$. For $x = 1$: $p(1) = (1)^{2} - 3(1) + 2 = 1 - 3 + 2 = 0$.
$2$. For $x = 2$: $p(2) = (2)^{2} - 3(2) + 2 = 4 - 6 + 2 = 0$.
$3$. For $x = 4$: $p(4) = (4)^{2} - 3(4) + 2 = 16 - 12 + 2 = 6$.
Thus,the values are $p(1) = 0$,$p(2) = 0$,and $p(4) = 6$.
$1$. For $x = 1$: $p(1) = (1)^{2} - 3(1) + 2 = 1 - 3 + 2 = 0$.
$2$. For $x = 2$: $p(2) = (2)^{2} - 3(2) + 2 = 4 - 6 + 2 = 0$.
$3$. For $x = 4$: $p(4) = (4)^{2} - 3(4) + 2 = 16 - 12 + 2 = 6$.
Thus,the values are $p(1) = 0$,$p(2) = 0$,and $p(4) = 6$.
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