Find a polynomial with degree n=3, with zeros of 1 and 5i. Where f(-1) = -104

Question

This is as far as I got

(x-1)(x+5i)(x-5i)

(x-1)(x^2+25)

x^3-x^2+25x-25

f(-1) = (-1)^3 - (-1)^2 + 25(-1) - 25

= -1 - 1 - 25 - 25

= -52

this does not equal -104, but is exactly half, what am I missing?

Answer 1

You can add a coefficient onto any polynomial which gives the transformation of a vertical stretch, but doesn't change the zeros.

Your starting point is correct.

(x-1)(x+5i)(x-5i)

This is the full equation:

a(x-1)(x+5i)(x-5i) = y

The next step is to plug in x and y, and solve for the coefficient, a

a((-1) -1)((-1)+5i)((-1)-5i) = -104

a(-2)(26) = -104

a(-52) = -104

a = 2

Multiplying it all out you get

y =2x^3-2x^2+50x-50

Answer 2

f(x) = 2 (x+5i)(x-5i)(x-1)

The 2 does not change the zeros of the function.