Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: $p(x) = x^3 - 4x^2 + x + 6$,$g(x) = x - 3$.
(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$g(x) = x - 3$,so $a = 3$.
We calculate $p(3)$ by substituting $x = 3$ into $p(x) = x^3 - 4x^2 + x + 6$:
$p(3) = (3)^3 - 4(3)^2 + (3) + 6$
$p(3) = 27 - 4(9) + 3 + 6$
$p(3) = 27 - 36 + 3 + 6$
$p(3) = 36 - 36 = 0$
Since $p(3) = 0$,by the Factor Theorem,$g(x) = x - 3$ is a factor of $p(x) = x^3 - 4x^2 + x + 6$.
Here,$g(x) = x - 3$,so $a = 3$.
We calculate $p(3)$ by substituting $x = 3$ into $p(x) = x^3 - 4x^2 + x + 6$:
$p(3) = (3)^3 - 4(3)^2 + (3) + 6$
$p(3) = 27 - 4(9) + 3 + 6$
$p(3) = 27 - 36 + 3 + 6$
$p(3) = 36 - 36 = 0$
Since $p(3) = 0$,by the Factor Theorem,$g(x) = x - 3$ is a factor of $p(x) = x^3 - 4x^2 + x + 6$.
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