SOLUTION: Find the polynomial function of least degree with integral coefficients that has the given zeros 2,5/3,-5?

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Question 1164756: Find the polynomial function of least degree with integral coefficients that
has the given zeros 2,5/3,-5?
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Final problem for my Math 3 homwework:find the polynomial function of least degree with integral coefficients that has the given zeros 2,5/3,-5?
-----------------
---> %28x-2%29%2A%28x-5%2F3%29%2A%28x%2B5%29
Multiply by 3 to eliminate the fraction.

Answer by Edwin McCravy(20065) 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 for you
to use as a model for doing yours.

Find the polynomial function of least degree with integral coefficients that
has the given zeros 3,7/4,-6?
    x = 3;          x = 7/4;       x = -6

Get 0 on the right of each equation:

x - 3 = 0;    x - 7/4 = 0      x + 6 = 0

Clear the second one of fractions by multiplying it through by 3.

So we have:

x - 3 = 0;     4x - 7 = 0      x + 6 = 0

Multiply the three left sides together:

       (x - 3)(4x - 7)(x + 6) = 0

Multiply the first two together by FOIL:

 (4x² - 7x - 12x + 21)(x + 6) = 0
      (4x² - 19x + 21)(x + 6) = 0

Multiply every term in the first parentheses by every term in the secon4
parentheses:

  4x³ + 24x² - 19x² - 114x + 21x + 126 = 0
                  4x³ + 5x² - 93x + 50 = 0

We want the polynomial that when set = 0, has those three solutions.  So
that polynomial is the left side:

          f(x) =  4x³ + 5x² - 93x + 50

Now go do yours the exact same way.

Edwin