6、设p(x)=x^4+ax^3+bx^2+cx+d,a,b为常数,且p(1)=1993,p(2)=3986,p(3)=5979,求1/4[p(11)+P(-7)]的值.

Question

6、设p(x)=x^4+ax^3+bx^2+cx+d,a,b为常数,且p(1)=1993,p(2)=3986,p(3)=5979,求1/4[p(11)+P(-7)]的值.

Answer 1

设q(x)=p(x) - 1993x,则由已知可得q(1)=q(2)=q(3)=0,即x=1,2,3都是四次多项式q(x)的零点,设q(x)第四个零点为y,则q(x)=(x-1)(x-2)(x-3)(x-y) = p(x) - 1993x ,所以
p(x) = (x-1)(x-2)(x-3)(x-y) +1993x,
所以1/4[p(11) + p(-7)] = 1/4[720(11-y) + 1993*11 + 720(7+y) + 1993*(-7)] = 5233.