Michael J. answered • 03/07/17

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Your zeros are

x

_{1}= 3x

_{2}= -4ix

_{3}= 4iAnd since you have a 4th degree polynomial, you need more root to have 4 zeros. Supposed the first root has a multiplicity of 2. This is because conjugate complex roots are associated with 2nd degree polynomials. -4i is the conjugate of 4i.

f(x) = C(x - 3)

^{2}(x + 4i)(x - 4i)where C is a constant. This C will be found using the condition f(1)=-234

f(x) = C(x - 3)

^{2}(x^{2}+ 16)Now we use the initial value condition to solve for C.

-234 = C(1 - 3)

^{2}(1^{2}+ 16)-234 = C(4)(17)

-234 = 68C

-234/68 = C

**f(x) = (-234/68)(x - 3)(x - 3)(x**

^{2}+ 16)
Mark M.

_{4}= 0?03/07/17