Find the polynomial having roots at 0, 2i and 3 - i, and passing through (1, –5).

A polynomial  which has the complex roots 2i and 3-i, must
have also the roots which are the conjugates of 2i and
3-i.

Therefore the give polynomial has the a ral root 0,
and complex roots 2i and -2i, 3-i and 3+i.

If a, b, c, are
the roots of a polynomial y(x), then y(x) =
k(x-a)(x-b)(x-c).

Therefore y(x) =
k(x-0)(x-2i)(9x-2i){x+3-i)(x-(3+i)) is the polynomial with roots 0, 2i, -2i, 3-i and
3+i.

y(x) =
kx(x^2+4)((x-3)^2+1).

y(x) = kx(x^2+4)(x^2-6x+10). This
polynomial should pass through (1-5).

Therefore
substituting (x ,y) = (1, -5) in y(x) = kx(x^2+2)(x^2-6x+10), we
get:

-5 =
k(1)(1+4)((1-6+10).

-5 =
k*25.

Therefore k = -5/25=
-1/5.

Therefore the required polynomial which has roots, 0,
2i, and 3-i and passes through the point (1,-5) is given
by:

 y(x) =
(-1/5)x{x^2+4)(x^2-6x+10).