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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Updated on: 10-Oct-2022

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