ANSWER
[tex]\text{ x}^3\text{ - 3x}^2\text{ + 3x - 5}[/tex]
EXPLANATION
Given that;
The roots of the polynomial function are zeros 1, 2i - 1
Recall that, since 2i - 1 is a root, then 2i + 1 is also a root
[tex]\text{ \lparen x - 1\rparen }\lbrack x\text{ - \lparen2i - 1\rparen}\rbrack\text{ }\lbrack x\text{ - \lparen2i + 1\rparen}\rbrack[/tex]
Expand the brackets
[tex]\begin{gathered} \text{ \lparen x - 1\rparen }\lbrack\text{\lparen x - 1\rparen}^2\text{ -\lparen2i\rparen}^2\rbrack \\ \text{ \lparen x - 1\rparen}^\text{ }\lbrack(x\text{ - 1\rparen}^2\text{ - 4i}^2\rbrack \\ \text{ \lparen x - 1\rparen }\lbrack\text{ x}^2\text{ - 2x + 1\rparen- 4i}^2 \\ \text{ Recall, }\sqrt{-1}\text{ = i} \\ \text{ Hence, i}^2\text{ = -1} \\ \text{ \lparen x - 1\rparen\lparen x}^2\text{ - 2x + 1 + 4\rparen } \\ \text{ \lparen x - 1\rparen\lparen x}^2\text{ - 2x + 5\rparen} \\ \text{ x}^3\text{ - 2x}^2\text{ + 5x - x}^2\text{ - 2x - 5} \\ \text{ x}^3\text{ - 3x}^2\text{ + 3x - 5} \end{gathered}[/tex]
