The function

Solution:

Here,

$x^{3}+x^{2}-16 x+20$

$=x^{3}-2 x^{2}+3 x^{2}-6 x-10 x+20$

$=x^{2}(x-2)+3 x(x-2)-10(x-2)$

$=(x-2)\left(x^{2}+3 x-10\right)$

$=(x-2)(x-2)(x+5)$

$=(x-2)^{2}(x+5)$

So, the given function can be rewritten as

$f(x)=\frac{(x-2)^{2}(x+5)}{x-2}$

$\Rightarrow f(x)=(x-2)(x+5)$

If $f(x)$ is continuous at $x=2$, then

$\lim _{x \rightarrow 2} f(x)=f(2)$

$\Rightarrow \lim _{x \rightarrow 2}(x-2)(x+5)=f(2)$

$\Rightarrow f(2)=0$

Hence, in order to make $f(x)$ continuous at $x=2, f(2)$ should be defined as $0 .$