Question 71793
Find the polynomial f(x) of degree 3 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)
Now find "a".
f(0)= a(0-1)(0-2)(0-4)=-16
a(-1)(-2)(-4)=-16
 -8a=-16
a=2
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Therefore f(x)=2(x-1)(x-2)(x-4)

Cheers,
Stan H.