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
