Giải thích các bước giải:
Để P(x) chia hết $(2x-1)^2\to P(\dfrac 12)=0$
$\to \dfrac{1}{2}+(m+n)\dfrac{1}{4}-(m+4n)\dfrac{1}{2}+2=0$
$\to m=10-7n$
$\to P(x)=4x^3+(10-6n)x^2-(10-3n)x+2$
$\to P(x)=2(x^2+3x-1)(2x-1)-3nx(2x-1)$
$\to P(x)=(2x-1)(2(x^2+3x-1)-3nx)$
$\to f(x)=2(x^2+3x-1)-3nx$
$\to f(\dfrac 12)=0\to 2(\dfrac 14+3.\dfrac 12+1)-\dfrac 32n =0$
$\to n=\dfrac{19}3\to m=10-7.\dfrac{19}{3}=\dfrac{-103}{3}$