Đặt \(f\left(x\right)=\left(x^2+1\right)\left(x+1\right)\cdot Q\left(x\right)+a\cdot x^2+bx+c\) , với \(ax^2+bx+c\) là dư sau khi chia f(x) chia \(\left(x+1\right)\left(x^2+1\right)\)
=>\(f\left(x\right)=\left(x^2+1\right)\left(x+1\right)\cdot Q\left(x\right)+a\left(x^2-1\right)+a+b\left(x+1\right)-b+c\)
=>\(f\left(x\right)=\left(x+1\right)\left\lbrack\left(x^2+1\right)\cdot Q\left(x\right)+a\left(x-1\right)+b\right\rbrack+a-b+c\)
f(x) chia x+1 dư 5 nên a-b+c=5
Ta có: \(f\left(x\right)=\left(x^2+1\right)\left(x+1\right)\cdot Q\left(x\right)+a\cdot x^2+bx+c\)
\(=\left(x^2+1\right)\left(x+1\right)\cdot Q\left(x\right)+a\cdot x^2+\frac32\cdot ax+bx-\frac32ax+c\)
\(=\left(x^2+1\right)\left(x+1\right)\cdot Q\left(x\right)+\frac12\cdot a\cdot x\cdot\left(2x+3\right)+x\left(b-\frac32a\right)+c\)
f(x) chia x^2+1 thì dư 2x+3 nên b-3/2a=2 và c=3
a-b+c=5
=>a-b=5-c=5-3=2
=>a=b+2
\(b-\frac32a=2\)
=>\(b-\frac32\left(b+2\right)=2\)
=>\(b-\frac32b-3=2\)
=>\(-\frac12b=5\)
=>b=-10
a=b+2=-10+2=-8
Vậy: Đa thức cần tìm là \(-8x^2-10x+3\)
