How do you long divide $\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4 + 4x^3 -8x + 2)/( 2x^2 - 3x + 1)$ ?

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How do you long divide $\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4 + 4x^3 -8x + 2)/( 2x^2 - 3x + 1)$ ?

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Answer 1 · 2016-03-27T20:31:10

Long divide coefficients to find:

$\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4+4x^3-8x+2)/(2x^2-3x+1)$

$= (2 x^{2} + 5 x + \frac{13}{2}) + \frac{\frac{13}{2} x - \frac{9}{2}}{2 x^{2} - 3 x + 1}$ $=(2x^2+5x+13/2) + (13/2x-9/2)/(2x^2-3x+1)$

Explanation:

I like to just long divide the coefficients, not forgetting to include $0$ $0$ 's for missing powers of $x$ $x$ . In addition, I'm not keen on fractions, so I will choose to multiply the numerator by $2$ $2$ first, then divide by $2$ $2$ at the end.

$2 \times (4 x^{4} + 4 x^{3} - 8 x + 2) = 8 x^{4} + 8 x^{3} + 0 x^{2} - 16 x + 4$ $2 xx (4x^4+4x^3-8x+2) = 8x^4+8x^3+0x^2-16x+4$

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This process is similar to long division of decimal numbers.

Write the dividend $(8, 8, 0, - 16, 4)$ $(8, 8, 0, -16, 4)$ under the bar and the divisor $(2, - 3, 1)$ $(2, -3, 1)$ to the left of the bar.

Choose the first term $4$ $color(blue)(4)$ of the quotient so that when multiplied by the divisor it matches the leading term of the dividend.

Write the product $(8, - 12, 4)$ $(8, -12, 4)$ under the dividend and subtract it to get a remainder. Bring down the next term $- 16$ $-16$ of the dividend alongside it to give the running remainder.

Choose the next term $10$ $color(blue)(10)$ of the quotient so that when multiplied by the divisor it matches the first term of the running remainder.

Write the product $(20, - 30, 10)$ $(20, -30, 10)$ under the running remainder and subtract it to get the new running remainder. Bring down the next term $4$ $4$ of the dividend alongside it.

Choose the final term $13$ $color(blue)(13)$ of the quotient as before, write down the product $(26, - 39, 13)$ $(26, -39, 13)$ and subtract it to give the final remainder $(13, - 9)$ $color(red)("("13, -9")")$ .

So:

$8 x^{4} + 8 x^{3} - 16 x + 4 = (2 x^{2} - 3 x + 1) (4 x^{2} + 10 x + 13) + (13 x - 9)$ $8x^4+8x^3-16x+4 = (2x^2-3x+1)(4x^2+10x+13)+(13x-9)$

Dividing by $2$ $2$ :

$4 x^{4} + 4 x^{3} - 8 x + 2$ $4x^4+4x^3-8x+2$

$= (2 x^{2} - 3 x + 1) (2 x^{2} + 5 x + \frac{13}{2}) + (\frac{13}{2} x - \frac{9}{2})$ $= (2x^2-3x+1)(2x^2+5x+13/2) + (13/2x-9/2)$

In other words:

$\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4+4x^3-8x+2)/(2x^2-3x+1)$

$= (2 x^{2} + 5 x + \frac{13}{2}) + \frac{\frac{13}{2} x - \frac{9}{2}}{2 x^{2} - 3 x + 1}$ $=(2x^2+5x+13/2) + (13/2x-9/2)/(2x^2-3x+1)$

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How do you long divide $\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4 + 4x^3 -8x + 2)/( 2x^2 - 3x + 1)$ ?

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Explanation:

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How do you long divide $\frac{4 x^{4} + 4 x^{3} - 8 x + 2}{2 x^{2} - 3 x + 1}$ $(4x^4 + 4x^3 -8x + 2)/( 2x^2 - 3x + 1)$ ?