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!!!
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