f(x)=(x2+6x+9)50−4x+3=(x+3)100−4x+3
Substituting r1,r2,⋯,r100 into f(x) = 0, we have
{(r1+3)100−4r1+3=0(r2+3)100−4r2+3=0⋯(r100+3)100−4r100+3=0
Moving all the terms to the right hand side and adding them up,
(r1+3)100+(r2+3)100+⋯+(r100+3)100=4(r1+r2+⋯+r100)−300
That means the required answer is 4(sum of roots) - 300.
By Vieta's formula, sum of roots=−coefficient of x99coefficient of x100.
Using binomial theorem, the coefficient of x100 is 1 and the coefficient of x99 is 300.
Therefore r1+r2+⋯+r100=−300.
Therefore (r1+3)100+(r2+3)100+⋯+(r100+3)100=4(−300)−300=−1500
.