Question

Need help with these two questions number 8 and number 9 please

Answer 1

`\text{ x}^3\text{ - 3x}^2\text{ + 3x - 5}`

Step-by-step 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

`\text{ \lparen x - 1\rparen }\lbrack x\text{ - \lparen2i - 1\rparen}\rbrack\text{ }\lbrack x\text{ - \lparen2i + 1\rparen}\rbrack`

Expand the brackets

`\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, }√(-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}`