SOLUTION: Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies the given condition. −4i, 4i, 2; f(1) = 51

Algebra ->  Inequalities -> SOLUTION: Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies the given condition. −4i, 4i, 2; f(1) = 51      Log On


   

Question 1171426: Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies
the given condition.
−4i, 4i, 2; f(1) = 51
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing your problem for you, I'll do one exactly like it that uses
exactly the same steps.  I'll do this one instead and you can use it as a
model to go by:

Find a polynomial f(x) of degree 3 that has the indicated zeros and satisfies
the given condition.−5i, 5i, 3; f(2) = 116
x = -5i; x = 5i; x = 3;

x + 5i = 0; x - 5i = 0; x - 3 = 0

Multiply left and right sides together

(x + 5i)(x - 5i)(x - 3) = 0

Multiply the first two parentheses using FOIL:

%28x%5E2-25i%5E2%29%28x-3%5E%22%22%29=0

Since i2 = -1 substitute that:

%28x%5E2-25%28-1%29%5E%22%22%29%28x-3%5E%22%22%29=0

%28x%5E2%2B25%5E%22%22%29%28x-3%5E%22%22%29=0

Use FOIL again:

x%5E3-3x%5E2%2B25x-75=0

So f(x) is a constant k times the left side of that:

%22f%28x%29%22=k%28x%5E3-3x%5E2%2B25x-75%29

Since f(2) = 116, we substitute x=2

%22f%282%29%22=k%282%5E3-3%282%29%5E2%2B25%282%29-75%29

116=k%282%5E3-3%282%29%5E2%2B25%282%29-75%29

116=k%288-3%284%29%2B50-75%5E%22%22%29

116=k%288-12%2B50-75%5E%22%22%29


116=k%28-29%29

%28116%29%2F%28-29%29=k

-4=k

%22f%28x%29%22=-4%28x%5E3-3x%5E2%2B25x-75%29
 
%22f%28x%29%22=-4x%5E3%2B12x%5E2-100x%2B300

Now do yours exactly the same way.

Edwin