How do you divide (x^4-2x^2+3)/(x-1)?

Dec 3, 2015

${x}^{3} + {x}^{2} - x - 1 + \frac{2}{x - 1}$

Explanation:

I know that in some countries, a different notation of the polynomial long division is being used. I would like to use the notation that I'm familiar with, and I hope that it will be no problem to re-write the division in your notation if needed!

$\textcolor{w h i t e}{\xi i} \left({x}^{4} \textcolor{w h i t e}{\times \times} - 2 {x}^{2} \textcolor{w h i t e}{\times \times \times} + 3\right) \div \left(x - 1\right) = {x}^{3} + {x}^{2} - x - 1$
$- \left({x}^{4} - {x}^{3}\right)$
$\textcolor{w h i t e}{\xi i} \frac{\textcolor{w h i t e}{\times \times \times}}{}$
$\textcolor{w h i t e}{\times \times \times} {x}^{3} - 2 {x}^{2}$
$\textcolor{w h i t e}{\times \xi i} - \left({x}^{3} - \textcolor{w h i t e}{x} {x}^{2}\right)$
$\textcolor{w h i t e}{\times \times \xi i} \frac{\textcolor{w h i t e}{\times \times \times x}}{}$
$\textcolor{w h i t e}{\times \times \times \times} - {x}^{2} + 0 \cdot x$
$\textcolor{w h i t e}{\times \times \xi i} - \left(- {x}^{2} + \textcolor{w h i t e}{\times} x \textcolor{w h i t e}{i}\right)$
$\textcolor{w h i t e}{\times \times \times \xi i} \frac{\textcolor{w h i t e}{\times \times \times \times \times}}{}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times} - x \textcolor{w h i t e}{x} + 3$
$\textcolor{w h i t e}{\times \times \times \times \times \times} - \left(- x \textcolor{w h i t e}{x} + 1\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times i i} \frac{\textcolor{w h i t e}{\times \times \times x}}{}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times i i} 2$

So, you have a remainder $2$, and in total, your solution is

$\frac{{x}^{4} - 2 {x}^{2} + 3}{x - 1} = {x}^{3} + {x}^{2} - x - 1 + \frac{2}{x - 1}$