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Find the polynominal f(X) of degree three
that has zeroes at 1, 2, and 4 such that
f(0) = -16.
To have zero 1 it must have factor (x - 1)
To have zero 2 it must have factor (x - 2)
To have zero 4 it must have factor (x - 4)
It may have any non-zero constant factor,
say k
Then
f(x) = k(x - 1)(x - 2)(x - 4)
f(0) = k(0 - 1)(0 - 2)(0 - 4)
f(0) = k(-1)(-2)(-4)
f(0) = -8k
Since we are given f(0) = -16
-16 = -8k
2 = k
So
f(x) = k(x - 1)(x - 2)(x - 4)
f(x) = 2(x - 1)(x - 2)(x - 4)
f(x) = 2(x - 1)(x² - 6x + 8)
f(x) = 2(x³ - 6x² + 8x - x² + 6x - 8)
f(x) = 2(x³ - 7x² + 14x - 8)
f(x) = 2x³ - 14x² + 28x - 16
Edwin McCravy
