Divide $x^{5} - x^{4} + 3^{3} + 4 x^{2} - 3 x - 3$ by $x^{2} + 1$

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Solution

Divide $x^{5} - x^{4} + 3^{3} + 4 x^{2} - 3 x - 3 \div (x^{3} + 1)$
Use long division method

$∴ Q = x^{3} - x^{2} + 2 x + 5 R = - 5 x - 8$
you cannot eliminate x. hence division is not possible or cannot have proper factorisation

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Divide $(x^{2} + 3) (x^{2} - 3) + 9$ by $x^{4}$ [1 MARK]

The degree of the polynomial obtained when $8 - 6 x + x^{2} - 7 x^{3} + x^{5}$ is subtracted from $x^{4} - 6 x^{3} + x^{2} - 3 x + 1$ is:

x^{4} - 4 x^{2} + 12 x + 9

x^{2} + 2 x - 3

Subtract $(8 - 6 x + x^{2} - 7 x^{3} + x^{5})$ from $(x^{4} - 6 x^{3} + x^{2} - 3 x + 1)$ .

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

\sqrt{x_{1} - 1} + 2 \sqrt{x_{2} - 4} + 3 \sqrt{x_{3} - 9} + 4 \sqrt{x_{4} - 16} + 5 \sqrt{x_{5} - 25} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{2}

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