Question

p(x)=x4+ax3+bx2+cx+d such that p(1)〓1 p(2)〓8 p(3)〓27 p(4)〓64.then find p(10)/2012

Answer 1

p(1)=1=1+a+b+c+d, so a+b+c+d=0.

p(2)=8=16+8a+4b+2c+d.

p(3)=27=81+27a+9b+3c+d.

p(4)=64=256+64a+16b+4c+d.

(A): p(2)=8=16+7a+3b+c

(B): p(3)-p(2): 19=65+19a+5b+c

(C): (B)-(A): 11=49+12a+2b, 12a+2b=-38, 6a+b=-19.

(D): p(4)-p(3): 37=175+37a+7b+c.

(E): (D)-(B): 18=110+18a+2b, 18a+2b=-92, 9a+b=-46.

(F): (E)-(C): 3a=-46+19=-27, a=-9.

b=-46-9a=-46+81=35.

c=-8-7a-3b=-8+63-105=-50.

d=-(a+b+c)=-(-9+35-50)=24.

Solution: (a,b,c,d)=(-9,35,-50,24) so p(x)=x^4-9x^3+35x^2-50x+24.

p(10)/2012=4024/2012=2.