Question

Given a polynomial and one of its factors, find the remaining factors of the polynomial.

x^3-9x^2+27x-27; x-3

Answer 1

(x-3)^3

Explanation:

`= x^3-9x^2+27x-27 \\ =(x − 3) (x² + 3x + 9) − 9x (x - 3) \\ =(x − 3) · (x² + 3x + 9 − 9x) \\ =(x-3)·(x²-6x+9) \\ =(x − 3) · (x − 3)^2 \\ = {(x - 3)}^(3)`

Or Apply cube of difference rule

`{x}^(3) - 3 {x}^(2) y + 3x {y}^(2) - {y}^(3)`

Answer 2

`(x-3)^3`

Explanation:

Given polynomial:

`x^3-9x^2+27x-27`

Given (x - 3) is a factor of the polynomial, divide the polynomial by the factor using long division to find the other factor:

`\large \begin{array}{r}x^2-6x+9\phantom{)}\\x-3{\overline{\smash{\big)}\,x^3-9x^2+27x-27\phantom{)}}}\\{-~\phantom{(}\underline{(x^3-3x^2)\phantom{-b0000000)}}\\-6x^2+27x-27\phantom{)}\\-~\phantom{()}\underline{(-6x^2+18x)\phantom{0000.}}\\9x-27\phantom{)}\\-~\phantom{()}\underline{(9x-27)\phantom{}}\\0\phantom{)}\end{array}`

Therefore:

`x^3-9x^2+27x-27=(x-3)(x^2-6x+9)`

Factor (x² - 6x + 9):

`\implies x^2-6x+9`

`\implies x^2-3x-3x+9`

`\implies x(x-3)-3(x-3)`

`\implies (x-3)(x-3)`

Therefore, the fully factored polynomial is:

`\implies x^3-9x^2+27x-27=(x-3)(x-3)(x-3)`

`\implies x^3-9x^2+27x-27=(x-3)^3`