Answer: [tex]\Large\boxed{h(x)=x-5}[/tex]
Step-by-step explanation:
[tex]\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppp\displaystyle\ \Large \boldsymbol{} h(x)=\frac{f(x)}{g(x)} =? \\\\\\f(x)=x^2-8x+15\\\\\\x^2-8x+\underbrace{15+1-1}_{15}=0 \\\\\\\underbrace{x^2-8x+16}_{(x-4)^2}-1=0 \\\\\\(x-4)^2-1\Longrightarrow f(x)=(x-5)(x-3)[/tex] [tex]\displaystyle\ \Large \boldsymbol{} h(x)=\frac{(x-3)(x-5)}{(x-3)}=\boxed{x-5}[/tex]
