Check whether $p(x)$ is a multiple of $g(x)$ or not :
$p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2$
Check whether $p(x)$ is a multiple of $g(x)$ or not :
$p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2$
$p ( x )$ will be a multiple $g ( x )$ if $g ( x )$ divides $p ( x )$
Now, $g(x)=x-2$ gives $x=2$
Remainder $=p(2)=(2)^{3}-5(2)^{2}+4(2)-3$
$=8-5(4)+8-3=8-20+8-3$
$=-7$
since remainder $\neq 0,$
So $p ( x )$ is not a multiple of $g ( x )$
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