Tìm số dư trong phép chia f(x) cho g(x) biết f(x)=(x+1)(x+3)(x+4)(x+6)+7 g(x)=x^2+7x+5

Ta có:

$f(x) = (x+1)(x+3)(x+4)(x+6) + 7$

$=[(x+1)(x+6)][(x+3)(x+4)] + 7$

$= (x^2 + 7x + 6)(x^2 + 7x + 12) + 7$

$= (g(x) + 1)(g(x) + 7) + 7$

$= g^2(x) + 8g(x) + 14$

$= g(x)(g(x) + 8) + 14$

$\Rightarrow f(x) = g(x).(x^2 + 7x + 13) + 14$

Vậy $f(x)$ chia $g(x)$ dư $14$