Show that $x+3$ is a factor of $69+11x-x^2+x^3$.
(N/A) Let $p(x) = 69 + 11x - x^2 + x^3$ and $g(x) = x + 3$.
According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$ where $g(a) = 0$.
Set $g(x) = x + 3 = 0$,which gives $x = -3$.
Now,calculate $p(-3)$:
$p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$
$p(-3) = 69 - 33 - 9 - 27$
$p(-3) = 69 - 69 = 0$.
Since $p(-3) = 0$,by the Factor Theorem,$x + 3$ is a factor of $69 + 11x - x^2 + x^3$.
According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$ where $g(a) = 0$.
Set $g(x) = x + 3 = 0$,which gives $x = -3$.
Now,calculate $p(-3)$:
$p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$
$p(-3) = 69 - 33 - 9 - 27$
$p(-3) = 69 - 69 = 0$.
Since $p(-3) = 0$,by the Factor Theorem,$x + 3$ is a factor of $69 + 11x - x^2 + x^3$.
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