How do you divide $\frac{x^{4} - 2 x^{2} + 3}{x - 1}$ $(x^4-2x^2+3)/(x-1)$ ?

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How do you divide $\frac{x^{4} - 2 x^{2} + 3}{x - 1}$ $(x^4-2x^2+3)/(x-1)$ ?

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Answer 1 · 2015-12-03T11:42:09

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

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!

$ξ i (x^{4} \times \times - 2 x^{2} \times \times \times + 3) \div (x - 1) = x^{3} + x^{2} - x - 1$ $color(white)(xii) (x^4 color(white)(xxxx) - 2x^2 color(white)(xxxxxx) + 3) -: (x- 1) = x^3 + x^2 -x - 1$
$- (x^{4} - x^{3})$ $-(x^4 - x^3)$
$ξ i \frac{\times \times \times}{}$ $color(white)(xii)(color(white)(xxxxxx))/()$
$\times \times \times x^{3} - 2 x^{2}$ $color(white)(xxxxxx) x^3 - 2x^2$
$\times ξ i - (x^{3} - x x^{2})$ $color(white)(xxxii) -(x^3 - color(white)(x) x^2)$
$\times \times ξ i \frac{\times \times \times x}{}$ $color(white)(xxxxxii)(color(white)(xxxxxxx))/()$
$\times \times \times \times - x^{2} + 0 \cdot x$ $color(white)(xxxxxxxx) - x^2 + 0 * x$
$\times \times ξ i - (- x^{2} + \times x i)$ $color(white)(xxxxxii) - (- x^2 + color(white)(xx) x color(white)(i))$
$\times \times \times ξ i \frac{\times \times \times \times \times}{}$ $color(white)(xxxxxxxii)(color(white)(xxxxxxxxxx))/()$
$\times \times \times \times \times \times \times - x x + 3$ $color(white)(xxxxxxxxxxxxxx) -x color(white)(x) + 3$
$\times \times \times \times \times \times - (- x x + 1)$ $color(white)(xxxxxxxxxxxx) - (-x color(white)(x)+1)$
$\times \times \times \times \times \times \times i i \frac{\times \times \times x}{}$ $color(white)(xxxxxxxxxxxxxxii)(color(white)(xxxxxxx))/()$
$\times \times \times \times \times \times \times \times \times \times i i 2$ $color(white)(xxxxxxxxxxxxxxxxxxxxii) 2$

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

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

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How do you divide $\frac{x^{4} - 2 x^{2} + 3}{x - 1}$ $(x^4-2x^2+3)/(x-1)$ ?

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