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

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

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Answer 1 · 2015-12-03T09:39:19

$\frac{2 x^{3} - 3 x^{2} + 3 x - 4}{x - 2} = 2 x^{2} + x + 5 + \frac{6}{x - 2}$ $(2x^3 - 3x^2+3x-4)/(x-2) = 2x^2 + x + 5 + (6)/(x-2)$

Explanation:

I know that in some countries, the long division of polynoms is being written in a different way. I will use the one that is familiar for me though and hope that you can convert it into your notation easily. :-)

Let me explain to you how to do the long division - and if you already know, you can skip to the end of the answer to see the whole division there.

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1) First, check that the the terms in your numerator and denominator are ordered by the power of $x$ $x$ - this is already the case for you.

2) Now, take the first term - the one with the biggest power - from the numerator and divide it by the first term - the one with the biggest power - from the denominator.

In your case, it's $2 x^{3} \div x = 2 x^{2}$ $2x^3 -: x = 2x^2$ , so you get:

$ξ i (2 x^{3} - 3 x^{2} x + 3 x - 4) \div (x - 2) = 2 x^{2}$ $color(white)(xii)(color(red)(2x^3) - 3x^2 color(white)(x)+ 3x - 4) -: (color(red)(x)-2) = color(red)(2x^2)$

3) As next, you need to backwards multiply your new result ( $2 x^{2}$ $2x^2$ ) with the denominator, so compute $2 x^{2} \cdot (x - 2) = 2 x^{3} - 4 x^{2}$ $2x^2 * (x-2) = 2x^3 - 4x^2$ and subtract it from your numerator:

$ξ i (2 x^{3} - 3 x^{2} x + 3 x - 4) \div (x - 2) = 2 x^{2}$ $color(white)(xii)(2x^3 - 3x^2 color(white)(x) + color(grey)(3x) - 4) -: (color(blue)(x-2)) = color(blue)(2x^2)$
$- (2 x^{3} - 4 x^{2})$ $-(color(blue)(2x^3 - 4x^2))$
$x \frac{\times \times \times \times}{}$ $color(white)(x)(color(white)(xxxxxxxx))/()$
$\times \times \times x x^{2} + 3 x$ $color(white)(xxxxxxx) x^2 + color(grey)(3x)$

4) As expected, you don't have a $x^{3}$ $x^3$ term anymore, and your new term with the highest power of $x$ $x$ in the numerator is $x^{2}$ $x^2$ , so you need to divide $x^{2}$ $x^2$ by $x$ $x$ :

$ξ i (2 x^{3} - 3 x^{2} x + 3 x - 4) \div (x - 2) = 2 x^{2} + x$ $color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x - 4) -: (color(red)(x)-2) = 2x^2 + color(red)(x)$
$- (2 x^{3} - 4 x^{2})$ $-(2x^3 - 4x^2)$
$x \frac{\times \times \times \times}{}$ $color(white)(x)(color(white)(xxxxxxxx))/()$
$\times \times \times x x^{2} + 3 x$ $color(white)(xxxxxxx) color(red)(x^2) + 3x$

5) Again, do the backward multiplication and subtract the result:

$ξ i (2 x^{3} - 3 x^{2} x + 3 x x - 4) \div (x - 2) = 2 x^{2} + x$ $color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) color(grey)(- 4)) -: (color(blue)(x-2)) = 2x^2 + color(blue)(x)$
$- (2 x^{3} - 4 x^{2})$ $-(2x^3 - 4x^2)$
$x \frac{\times \times \times \times}{}$ $color(white)(x)(color(white)(xxxxxxxx))/()$
$\times \times \times x x^{2} + 3 x$ $color(white)(xxxxxxx) x^2 + 3x$
$\times ξ i x - (x^{2} - 2 x)$ $color(white)(xxxiix) -(color(blue)(x^2 - 2x))$
$\times \times \times \frac{\times \times \times}{}$ $color(white)(xxxxxx)(color(white)(xxxxxx))/()$
$\times \times \times \times \times x 5 x x - 4$ $color(white)(xxxxxxxxxxx) 5x color(white)(x)color(grey)(- 4)$

6) As next, divide $5 x$ $5x$ by $x$ $x$ ...

$ξ i (2 x^{3} - 3 x^{2} x + 3 x x - 4) \div (x - 2) = 2 x^{2} + x + 5$ $color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) - 4) -: (color(red)(x)-2) = 2x^2 + x + color(red)(5)$
$- (2 x^{3} - 4 x^{2})$ $-(2x^3 - 4x^2)$
$x \frac{\times \times \times \times}{}$ $color(white)(x)(color(white)(xxxxxxxx))/()$
$\times \times \times x x^{2} + 3 x$ $color(white)(xxxxxxx) x^2 + 3x$
$\times ξ i x - (x^{2} - 2 x)$ $color(white)(xxxiix) -(x^2 - 2x)$
$\times \times \times \frac{\times \times \times}{}$ $color(white)(xxxxxx)(color(white)(xxxxxx))/()$
$\times \times \times \times \times x 5 x x - 4$ $color(white)(xxxxxxxxxxx) color(red)(5x) color(white)(x)- 4$

7)... and perform the backward multiplication and subtraction...

$ξ i (2 x^{3} - 3 x^{2} x + 3 x x - 4) \div (x - 2) = 2 x^{2} + x + 5$ $color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) - 4) -: (color(blue)(x-2)) = 2x^2 + x + color(blue)(5)$
$- (2 x^{3} - 4 x^{2})$ $-(2x^3 - 4x^2)$
$x \frac{\times \times \times \times}{}$ $color(white)(x)(color(white)(xxxxxxxx))/()$
$\times \times \times x x^{2} + 3 x$ $color(white)(xxxxxxx) x^2 + 3x$
$\times ξ i x - (x^{2} - 2 x)$ $color(white)(xxxiix) -(x^2 - 2x)$
$\times \times \times \frac{\times \times \times}{}$ $color(white)(xxxxxx)(color(white)(xxxxxx))/()$
$\times \times \times \times \times x 5 x x - 4$ $color(white)(xxxxxxxxxxx) 5x color(white)(x)- 4$
$\times \times \times \times i - (5 x x - 10)$ $color(white)(xxxxxxxxi) -(color(blue)(5x color(white)(x)- 10))$
$\times \times \times \times \times x \frac{\times \times \times x}{}$ $color(white)(xxxxxxxxxxx)(color(white)(xxxxxxx))/()$
$\times \times \times \times \times \times \times \times i 6$ $color(white)(xxxxxxxxxxxxxxxxi) 6$

8) At this point, as you can't divide $6 \div x$ $6 -: x$ anymore, you have finished your division and have the remainder $6$ $6$ at the end:

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Solution:

$\frac{2 x^{3} - 3 x^{2} + 3 x - 4}{x - 2} = 2 x^{2} + x + 5 + \frac{6}{x - 2}$ $(2x^3 - 3x^2+3x-4)/(x-2) = 2x^2 + x + 5 + (6)/(x-2)$

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