Verified answer

f(x) = x(x+3)(x^2-10x+26)

x = 5 ± i

=> x - 5 = ±i

=> (x - 5)² = -1

=> (x - 5)² + 1 = 0

=> x² -10x + 25 + 1 = 0

=> x² -10x + 26 = 0

x = -3

=> x + 3 = 0

and of course x = 0

=> f(x) = ax(x + 3)(x² -10x + 26)

for any value of a f(x) will have the 4 listed roots.

Just use the zero product property in reverse. Multiply together binomials that have those numbers as roots:

f(x) = (x - 0) (x + 3) [x - (5 + i)] [x - (5 - i)]

The zero-product property says that f(x) is zero if and only if one of those factors is zero, so f has the given roots 0, -3, 5+i, and 5-i, and no other roots. The last two factors can be rewritten as:

[(x - 5) - i] [(x - 5) + i] = (x - 5)^2 - i^2 = x^2 - 10x + 25 + 1

That's how you get f(x) = x(x+3)(x^2 - 10x + 26)

Real solutions are 0 and -3

So x(x+3) are factors

Complex solutions

(x-5)^2 = -1

x^2-10x+26 = 0

f(x) = x(x+3)(x^2-10x+26)

f(x) = x^4-7x^3-4x^2+78x