Question 62537
Find the polynomial f(x) of degree three that has zeroes at 1,2 and 4 such that f(0)= -16.

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f(x)=a(x-1)(x-2)(x-4)
Then f(0)=a(-1)(-2)(-4)=-8a
So, -8a=-16
a=2
Therefore f(x)=2(x-1)(x-2)(x-4)
Cheers,
Stan H.