SOLUTION: FIND THE POLYNOMIAL f(x) OF DEGREE THREE THAT HAS ZEROES AT 1,2, AND 4 SUCH THAT f(0)=-16.

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Question 62976: FIND THE POLYNOMIAL f(x) OF DEGREE THREE THAT HAS ZEROES AT 1,2, AND 4 SUCH THAT f(0)=-16.
Found 2 solutions by uma, Nate:
Answer by uma(370)   (Show Source): You can put this solution on YOUR website!
Given that the polynomial has zeroes at 1, 2 and 4.
So the required polynomial will be :
a(x-1)(x-2)(x-4)
==> a(x-1)(x^2 - 2x - 4x + 8)
= a(x-1)(x^2 - 6x + 8)
= ax^3 - 6ax^2 + 8ax - ax^2 + 6ax - 8a
= ax^3 - 7ax^2 + 14ax - 8a
f(0) = - 8a
==> - 8a = - 16
==> a = 2
Thus f(x) = 2x^3 - 14x^2 + 28x - 16
is the required polynomial.
Good Luck!!!

Answer by Nate(3500)   (Show Source): You can put this solution on YOUR website!
(x - 1)(x - 2)(x - 4)
(x^2 - 3x + 2)(x - 4)
x^3 - 3x^2 + 2x - 4x^2 + 12x - 8
x^3 -7x^2 + 14x - 8 = f(x)

All values for x is zero ... so f(0) = - 8
2*f(0) = -16
2x^3 - 14x^2 + 28x - 16 = f(x)

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