SOLUTION: Find the polynmial f(x) of degree three that has zeroes at 1, 2 and 4 such that f(0)=-16. <p> I think it should be f(x)=d(x-1)(x-2)(x-4) so d(0-1)(0-2)(0-4) = f(0) = -16 So -6d

Algebra ->  Rational-functions -> SOLUTION: Find the polynmial f(x) of degree three that has zeroes at 1, 2 and 4 such that f(0)=-16. <p> I think it should be f(x)=d(x-1)(x-2)(x-4) so d(0-1)(0-2)(0-4) = f(0) = -16 So -6d       Log On


   

Question 16798: Find the polynmial f(x) of degree three that has zeroes at 1, 2 and 4 such that f(0)=-16.


I think it should be f(x)=d(x-1)(x-2)(x-4) so d(0-1)(0-2)(0-4) = f(0) = -16
So -6d = -16
d = 8/3


I'm not sure if I'm right so far and not when I plugged things in to the equation above I didn't get anything remotely right. Please help.
Thanks
Answer by rapaljer(4671) About Me  (Show Source):

You can put this solution on YOUR website!
It looks like you did it right. There is just one very small error in your solution:

I think it should be f(x)=d(x-1)(x-2)(x-4) so d(0-1)(0-2)(0-4) = f(0) = -16
So -6d = -16
d = 8/3

How about -8d = -16??
d= 2!!