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

All polynomials of degree n that have zeros r1, r2, ..., rn 
are of the form

f(x) = k(x-r1)(x-r2)···(x-rn) 

where k can be any non-zero number.

So any polynomial of degree 3 that has zeros 1, 2, and 4 is 
of the form

f(x) = k(x-1)(x-2)(x-4)

where k can be any non-zero number.

But we also want f(0) to equal -16. So it has to be true that 
if we substitute 0 for x and then -16 for f(0), they should be 
equal.  So,

f(x) = k(x-1)(x-2)(x-4)
f(0) = k(0-1)(0-2)(0-4)
 -16 = -8k
   2 = k

So substitute 2 for k in 

f(x) = k(x-1)(x-2)(x-4)

to give

f(x) = 2(x-1)(x-2)(x-4)    

Multiply all that out and you'll get

f(x) = 2x³ - 14x² + 28x - 16 

That is choice (b).

Edwin