<PRE><B>Question 66312</B>
Find the polynominal f(x) of degree three that has zeros at 1,2 and 4 such that f(0)=-16.
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The 3 factors will be (x-1)(x-2)(x-4),
: 
FOIL (x-1)(x-2) = x^2 -3x + 2,
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multiply by (x-4) and you get: x^3 - 7x^2 + 14x - 8;
:
But wait, it says f(0) = -16, so last term has to be -16:
Multiply the equation by 2 (,it will not effect the zeros)
:
2x^3 - 14x^2 + 28x - 16 is the equation that satisfies all these requirements
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