Question:
If $p(x)=x^{3}-5 x^{2}+4 x-3$ and $g(x)=x-2$, show that $p(x)$ is not a multiple of $g(x)$.
Solution:
$p(x)=x^{3}-5 x^{2}+4 x-3$
$g(x)=x-2$
Putting x = 2 in p(x), we get
$p(2)=2^{3}-5 \times 2^{2}+4 \times 2-3=8-20+8-3=-7 \neq 0$
Therefore, by factor theorem, (x − 2) is not a factor of p(x).
Hence, p(x) is not a multiple of g(x).