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If $x^3+5x^2-kx+6$ is divided by $( x-2)$, the remainder is $0$. Find the value of $k$.
Given: If $x^3+5x^2-kx+6$ is divided by $( x-2)$, the remainder is $0$.
To do: To find the value of $k$.
Solution:
Let $f( x)=x^3+5x^2-kx+6$ and $g( x)=x-2$
$\because$, when $f( x)=x^3+5x^2-kx+6$ is divided by $g( x)=x-2$, the remainder is $0$.
Therefore, $g( x)$ is a factor of $f( x)$.
When $x-2=0\Rightarrow x=2$, put this value in $f( x)$.
$f( x)=2^3+5( 2)^2-k( 2)+6=0$
$\Rightarrow 8+20-2k+6=0$
$\Rightarrow -2k+34=0$
$\Rightarrow -2k=-34$
$\Rightarrow k=\frac{-34}{-2}$
$\Rightarrow k=17$
Thus, $k=17$.
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