X^3-3x^2+81x-243 rewrite the polynomial in the form (x - d)(x - e)(x + f), where d is a real number and e and f are complex numbers of the form bi.

The given polynomial is ⇒⇒⇒  f(x) = x³ - 3x² + 81x - 243

by factoring the absolute term (243) to find one of the factors of the the polynomial
∴ 243 = (1 * 243) or (3 * 81) or  (9 * 27)
check which of the numbers {1 , 3 , 9 , 27 , 81 , 243} make f(x)= 0

i have checked 3 and it makes polynomial = 0

i.e:  f(3) = 0  ⇒⇒ (x - 3) is one of the factors of f(x)
By using the reminder theorem ⇒⇒ see the attache figure

∴
And ⇒⇒ (x² + 81)  is a sum of two squares which can be factored using the complex numbers as following
x² + 81 = ( x + 9i ) ( x - 9i )

∴ f(x) = x³ - 3x² + 81x - 243 = (x - 3)(x + 9i)(x - 9i)