How do you divide $\frac{2 x^{4} - 3 x^{3} - 5 x^{2} + 17 x + 1}{x^{2} - x}$ $( 2x^4 - 3x^3 - 5x^2+17x+1 )/(x^2 - x )$ ?

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How do you divide $\frac{2 x^{4} - 3 x^{3} - 5 x^{2} + 17 x + 1}{x^{2} - x}$ $( 2x^4 - 3x^3 - 5x^2+17x+1 )/(x^2 - x )$ ?

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Answer 1 · 2018-01-06T06:51:57

I'd like to use long division here. If you're not familiar with this method, I recommend this guide

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$ $* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *$

$.$ $color(white)(.)$

$x^{2} - x ∣ -$ $color(white)(x^2-x | -)$ $2 x^{4} - 3 x^{3} .$ $color(white)(2x^4-3x^3.)$ $2 x^{2} - x - 6$ $2x^2-x-6$
$x^{2} - x$ $color(white)(x^2-x)$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$
$x^{2} - x ∣ ∣$ $x^2-x|$ $-$ $color(white)(-)$ $2 x^{4} - 3 x^{3} - 5 x^{2} + 17 x + 1$ $2x^4-3x^3-5x^2+17x+1$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(black)(-)$ $2 x^{4} - 2 x^{3}$ $2x^4-2x^3$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $. . . . . . . . . .$ $. . . . . . . . . .$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $2 x^{4}$ $color(white)(2x^4)$ $- 1 x^{3} - 5 x^{2}$ $-1x^3-5x^2$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $2 x^{4}$ $color(white)(2x^4)$ $-$ $color(black)(-)$ $- 1 x^{3} + 1 x^{2}$ $-1x^3+1x^2$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $2 x^{4}$ $color(white)(2x^4)$ $. . . . . . . . . .$ $. . . . . . . . . .$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $2 x^{4} - 1 x^{3}$ $color(white)(2x^4 -1x^3)$ $- 6 x^{2} + 17 x$ $-6x^2 +17x$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $2 x^{4} - 1 x^{3}$ $color(white)(2x^4 -1x^3)$ $-$ $color(black)(-)$ $- 6 x^{2} + 6 x$ $-6x^2 +6x$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $2 x^{4} - 6 x^{2}$ $color(white)(2x^4-6x^2)$ $. . . . . . . . . .$ $. . . . . . . . . .$
$x^{2} - x ∣ ∣$ $color(white)(x^2-x | )$ $-$ $color(white)(-)$ $2 x^{4} - 1 x^{3} - 6 x^{2} . .$ $color(white)(2x^4 -1x^3-6x^2. .)$ $11 x + 1$ $11x +1$

We can't keep dividing because the divisor ( $x^{2} - x$ $x^2-x$ ) has a greater power (exponent) than the remainder ( $11 x + 1$ $11x+1$ ) So, this is our equation: $2 x^{2} - x - 6 + \frac{11 x + 1}{x^{2} - x}$ $2x^2-x-6+(11x+1)/(x^2-x)$

Note, we tacked on the remainder, divided by the original equation. This is essential!

To double check our work, we can graph the original problem and our new equation and see if they are the same

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How do you divide $\frac{2 x^{4} - 3 x^{3} - 5 x^{2} + 17 x + 1}{x^{2} - x}$ $( 2x^4 - 3x^3 - 5x^2+17x+1 )/(x^2 - x )$ ?

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How do you divide 2x4−3x3−5x2+17x+1x2−x( 2x^4 - 3x^3 - 5x^2+17x+1 )/(x^2 - x )?

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How do you divide $\frac{2 x^{4} - 3 x^{3} - 5 x^{2} + 17 x + 1}{x^{2} - x}$ $( 2x^4 - 3x^3 - 5x^2+17x+1 )/(x^2 - x )$ ?