polynomial that satisfies

Question

Write a polynomial f(x) that satisfies the given conditions.

 

Polynomial of lowest degree with zeros of 5/3 (multiplicity 2) and - 1/4 (multiplicity 1) and with f(0) = -50

 

f(x) = 

Answer 1

Multiplicity means the number of times a zero repeats.  We have two zeros that are 5/3 and one zero that is -1/4.  A zero means that if we evaluate f(x), the result is zero.  So the function we will have so far is

 

f(x) =C (3x - 5)²(4x + 1)

 

f(x) = C (3x - 5)(3x - 5)(4x + 1)

 

 

Now, we use the initial value condition f(0)=-50 to solve for C.

 

-50 = C(-5)(-5)(1)

 

-50 = 25C

 

-2 = C

 

 

Now we plug in this value of C into the polynomial.

 

f(x) = -2(3x - 5)(3x - 5)(4x + 1)

 

 

Expand the polynomial.

 

f(x) = -2(9x² - 30x + 25) (4x + 1)

 

f(x) =  (-18x² + 60x - 50)(4x + 1)

 

f(x) = 4x(-18x² + 60x - 50) - 18x² + 60x - 50

 

f(x) = -72x³ + 240x² - 200x - 18x² + 60x - 50

 

f(x) = -72x³ - 222x² - 140x - 50

Answer 2

f(x) = a(x-5/3)(x-5/3)(x+1/4)

 

      = a(1/3)(3x-5)(1/3)(3x-5)(1/4)(4x+1)

 

      = (a/36)(9x²-30x+25)(4x+1)

 

      = (a/36)(36x³+9x²-120x²-30x+100x+25)

 

      = (a/36)(36x³-111x²+70x+25)

 

Since f(0) = -50, 25a/36 = -50

 

                           a = -72

 

So, f(x) = -2(36x³-111x²+70x+25)

 

     f(x) = -72x³+222x²-140x-50