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$( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$, then the values of $a$ and $b$ are respectively.
Given: $( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$.
To do: To find the value of $a$ and $b$.
Solution:
As given, $( x-2)$ is a common factor of $x^{3}-4 x^{2}+a x+b$ and $x^{3}-a x^{2}+b x+8$
Let $x-2=0$
$\Rightarrow x=2$, Put this value in $x^{3}-4 x^{2}+a x+b$.
$x^{3}-4 x^{2}+a x+b$
$\Rightarrow 2^3-4( 2)^2+a\times 2+b=0$
$\Rightarrow 8-16+2a+b=0$
$\Rightarrow 2a+b-8=0$
$\Rightarrow 2a+b=8$ ....... $( i)$
Again, put $x=2$ value in $x^{3}-a x^{2}+b x+8$
$\Rightarrow 2^3-a( 2)^2+b( 2)+8=0$
$\Rightarrow 8-2a+2b+8=0$
$\Rightarrow -2a+2b=-16$ .......$( ii)$
On adding $( i)$ and $( ii)$,
$2a+b-2a+2b=8-16$
$\Rightarrow 3b=-8$
$\Rightarrow b=-\frac{8}{3}$, put this value in $( i)$
$2a-\frac{8}{3}=8$
$\Rightarrow 2a=8+\frac{8}{3}$
$\Rightarrow 2a=\frac{32}{3}$
$\Rightarrow a=\frac{32}{3}\times\frac{1}{2}$
$\Rightarrow a=\frac{16}{3}$
Thus, $a=\frac{16}{3}$ and $b=-\frac{8}{3}$