Write $f ( x ) = x ^ { 3 } - 11 x ^ { 2 } + 18 x + 25$ in the form $f ( x ) = ( x - k ) q ( x ) + r$ when $k = 6 + \sqrt { 6 }$.
A) $f ( x ) = [ x + ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } + ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) \right] - 5$
B) $f ( x ) = [ x + ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } + ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) \right] + 5$
C) $f ( x ) = [ x - ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } + ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) \right] + 5$
D) $f ( x ) = [ x - ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } + ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) \right] - 5$
E) $f ( x ) = [ x + ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } - ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) \right] - 5$

Question

Quizplus · Accepted Answer

We use long division or synthetic division to divide $f(x)$ by $x-k$, which gives us a quotient of $q(x)=x^2+(k-11)x+(18-k)$ and a remainder of $r=5$. Plugging in $k=6+\sqrt{6}$, we get \begin{align*}
q(x) &= x^2+(6+\sqrt{6}-11)x+(18-(6+\sqrt{6})) \
&= x^2+(-5+\sqrt{6})x+(12-\sqrt{6}).
\end{align*} Thus, we have \begin{align*}
f ( x ) &= ( x - k ) q ( x ) + r \
&= ( x - (6+\sqrt{6})) (x^2+(-5+\sqrt{6})x+(12-\sqrt{6})) + 5 \
&= [ x - ( 6 + \sqrt { 6 } ) ] \left[ x ^ { 2 } + ( - 5 + \sqrt { 6 } ) x - ( 6 - \sqrt { 6 } ) ight] - 5,
\end{align*} which is exactly choice $\boxed{	extbf{(D)}}$.

Write f(x)=x3−11x2+18x+25f ( x ) = x ^ { 3 } - 11 x ^ { 2 } + 18 x + 25f(x)=x3−11x2+18x+25

Write $f ( x ) = x ^ { 3 } - 11 x ^ { 2 } + 18 x + 25$