Question

x^3-3x+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.

Answer 1

$x^3-3x^2+81x-243=x^2(x-3)+81(x-3)=(x-3)(x^2+81)=(*)\\\\x^2+81=0\ \ |-81\\\\x^2=-81\to x=\pm\sqrt{-81}\to x=-9i\ \vee\ x=9i\\\\(*)=(x-3)(x-9i)(x+9i)$


$Answer:\ x^3-3x^2+81x-243=(x-3)(x+9i)(x-9i)$

Answer 2

Answer:

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Step-by-step explanation: