Question

Tìm đa thức bậc 4 f(x) biết f(0)=0, f(1)=6, f(-1)=0, f(2)=36, f(-2)=0

Answer 1

Theo đề bài, ta có:

f(x) = ax⁴ + bx³ + cx² + dx + e

=> \(\left\{{}\begin{matrix}f\left(1\right)=6\\f\left(-1\right)=0\\f\left(2\right)=36\\f\left(-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a+b+c+d=6\\a-b+c-d=0\\16a+8b+4c+2d=36\\16a-8b+4c-2d=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\frac{1}{2}\\b=2\\c=\frac{5}{2}\\d=1\end{matrix}\right.\) => f(x) = \(\frac{1}{2}x^4+2x^3+\frac{5}{2}x^2+x\)