Zero's of polynomial functions

Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

n=3

3 and 5i are zeros;

f (2)=29

f(x)=

Answer 1

zeros are 3 5i and -5i

one point on the cubic is (2, 29)

y = a(x-3)(x-5i)(x+5i)

y = a(x-3)(x^2 +25)

29 = a(-1)(29)

a = -1

y = -(x-3)(x^2 +25)

y = -x^3 +3x^2 -25x +75

check the answers of 1 & only 1 real zero = 3

-(3)^3 +3(3^2) -25(3) +75 = 0, 3 is the only real zero

Answer 2

This is a cubic function and the 3 zeros are 3, 5i and -5i.

P(x) = a(x - 3)(x - 5i)(x + 5i)

P(x) = a(x-3)(x^2 + 25)

P(x) = a(x^3 - 3x^2 + 25x - 75)

Now find a we will use the additional given point.

29 = a(2 - 3)(2^2 + 25)

29 = a(-1)(29)

a = -1

P(x) = -1(x^3 - 3x2 + 25x - 75)

P(x) = -x^3 + 3x^2 - 25x + 75

Answer 3

f(x) = a(x - 3)(x - 5i)(x + 5i)

29 =- a(2 - 3)(2 - 5i)(2 + 5i)'

Can you solve for a and answer?