Question

Tìm đa thức bậc ba f(x) biết: \(f\left(x\right)+f\left(x+1\right)=4x^3+14x^2+16x+17\)

Answer 1

Đặt \(f\left(x\right)=ax^3+bx^2+cx+d\)

\(\Rightarrow f\left(x+1\right)=a\left(x+1\right)^2+b\left(x+1\right)^2+c\left(x+1\right)+d\)

\(\Rightarrow f\left(x+1\right)=ax^3+\left(3a+b\right)x^2+\left(3a+2b+c\right)x+a+b+c+d\)

\(\Rightarrow f\left(x\right)+f\left(x+1\right)=2ax^3+\left(3a+2b\right)x^2+\left(3a+2b+2c\right)x+a+b+c+2d\)

Đồng nhất hệ số ta được:

\(\left\{{}\begin{matrix}2a=4\\3a+2b=14\\3a+2b+2c=16\\a+b+c+2d=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=4\\c=1\\d=5\end{matrix}\right.\)

Vậy \(f\left(x\right)=2x^3+4x^2+x+5\)