SOLUTION: Find the polynomial f(x) of degree three that zeroes at 1, 2, and 4 such that f(0)=-16. Thank you!

Question 44650This question is from textbook
: Find the polynomial f(x) of degree three that zeroes at 1, 2, and 4 such that f(0)=-16.
Thank you! This question is from textbook

Found 2 solutions by adamchapman, Nate:
Answer by adamchapman(301) (Show Source): You can put this solution on YOUR website!
We know that:
,
,
.
And the Highest power of x is cubed
This step is pretty simple just put in brackets (x-a) where a is the value of x where f(x)=0 (i.e. f(a)=0):
(Don't worry, the value of "b" is about to become clearer).
We also know that when x=0, f(x)=-16. Try to find the value of f(0) with our previous equation:

so b=2.
Now we can write the function:

here's the graph of y=f(x):

I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam

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 Fill in zero for 'x' to get -8 .... -8x = -16 : x = 2
2x^3 - 14x^2 + 28x - 16 = f(x)

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