Find the polynomial f(x) of degree three that has zeros at 1,2, and 4
such that f(0)=-16. My answer is f(x)=x^3-7x^2+14x-16, could this
be right. thanks for checking my answer.
No, that couldn't be the right answer because
f(1) = (1)³ - 7(1)² + 14(1) - 16 = 1 - 7 + 14 - 16 = -8
But f(1) would have to be 0, not -8, if 1 were a zero.
1. If r is a zero of a polynomial then (x-r) is a factor of the
polynomial.
2. If a polynomial is multiplied by a constant k, it still has the
same zeros.
By 1, since 1 is a zero, (x-1) must be a factor of the polynomial.
By 1, since 2 is a zero, (x-2) must be a factor of the polynomial.
By 1, since 4 is a zero, (x-4) must be a factor of the polynomial.
By 2, we can assume the polynomial has factor k, so we have
f(x) = k(x-1)(x-2)(x-4)
Since f(0) = -16
f(0) = -16 = k(0-1)(x-2)(0-4) =
k(-1)(-2)(-4) =
-8k
since -16 = -8k, then k = 2, so
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
AnlytcPhil@aol.com
