Find $p(1)$,$p(2)$,and $p(4)$ for the polynomial $p(x) = x^3 + 9x^2 + 23x + 15$.
To find the values,substitute the given values of $x$ into the polynomial $p(x) = x^3 + 9x^2 + 23x + 15$:
$1$. For $x = 1$:
$p(1) = (1)^3 + 9(1)^2 + 23(1) + 15 = 1 + 9 + 23 + 15 = 48$.
$2$. For $x = 2$:
$p(2) = (2)^3 + 9(2)^2 + 23(2) + 15 = 8 + 9(4) + 46 + 15 = 8 + 36 + 46 + 15 = 105$.
$3$. For $x = 4$:
$p(4) = (4)^3 + 9(4)^2 + 23(4) + 15 = 64 + 9(16) + 92 + 15 = 64 + 144 + 92 + 15 = 315$.
$1$. For $x = 1$:
$p(1) = (1)^3 + 9(1)^2 + 23(1) + 15 = 1 + 9 + 23 + 15 = 48$.
$2$. For $x = 2$:
$p(2) = (2)^3 + 9(2)^2 + 23(2) + 15 = 8 + 9(4) + 46 + 15 = 8 + 36 + 46 + 15 = 105$.
$3$. For $x = 4$:
$p(4) = (4)^3 + 9(4)^2 + 23(4) + 15 = 64 + 9(16) + 92 + 15 = 64 + 144 + 92 + 15 = 315$.
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