Question

用秦九韶方法求多項式f(x)= x⁷-2x⁶+3x³-4x²+1在x=2時的函數值.

Answer 1

解：f(x)= x⁷-2x⁶+3x³-4x²+1=((((((x-2)x+0)x+0)x+3)x-4)x+0)x+1.
由內向外逐次計算:
v₀="1, " v₁=v₀x+a₆="1×2-2=0, " v₂=v₁x+a₅="0×2+0=0,"
v₃=v₂x+a₄="0×2+0=0," v₄=v₃x+a₃="0×2+3=3,"
v₅=v₄x+a₂="3×2-4=2," v₆=v₅x+a₁=2×2+0=4,
v₇=v₆x+a₀=4×2+1=9.
故f(2)=9
根據秦九韶算法的操作方法，先將多項式f(x)進行改寫，再逐步求值.