How do you long divide $\frac{5 x^{4} - 2 x^{3} - 7 x^{2} - 39}{x^{2} + 2 x - 4}$ $(5x^4-2x^3- 7x^2 -39) / (x^2 + 2x- 4)$ ?

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Answer 1 · 2015-06-06T23:44:29

The first multiplier is $5 x^{2}$ $color(red)(5x^2)$

$5 x^{2} (x^{2} + 2 x - 4) = 5 x^{4} + 10 x^{3} - 20 x^{2}$ $5x^2(x^2+2x-4) = 5x^4+10x^3-20x^2$

Subtract this from the original numerator to get a remainder:

$(5 x^{4} - 2 x^{3} - 7 x^{2} - 39) - (5 x^{4} + 10 x^{3} - 20 x^{2})$ $(5x^4-2x^3-7x^2-39)-(5x^4+10x^3-20x^2)$

$= - 12 x^{3} + 13 x^{2} - 39$ $=-12x^3+13x^2-39$

The second multiplier is $- 12 x$ $color(red)(-12x)$

$- 12 x (x^{2} + 2 x - 4) = - 12 x^{3} - 24 x^{2} + 48 x$ $-12x(x^2+2x-4) = -12x^3-24x^2+48x$

Subtract this from the remainder to get a new remainder:

$(- 12 x^{3} + 13 x^{2} - 39) - (- 12 x^{3} - 24 x^{2} + 48 x)$ $(-12x^3+13x^2-39) - (-12x^3-24x^2+48x)$

$= 37 x^{2} - 48 x - 39$ $=37x^2-48x-39$

The next multiplier is $37$ $color(red)(37)$

$= 37 (x^{2} + 2 x - 4) = 37 x^{2} + 74 x - 148$ $=37(x^2+2x-4) = 37x^2+74x-148$

Subtract this from the remainder to get a new remainder:

$(37 x^{2} - 48 x - 39) - (37 x^{2} + 74 x - 148)$ $(37x^2-48x-39) - (37x^2+74x-148)$

$= - 122 x - 187$ $=-122x-187$

Add the multiplier we have found together to get:

$\frac{5 x^{4} - 2 x^{3} - 7 x^{2} - 39}{x^{2} + 2 x - 4}$ $(5x^4-2x^3-7x^2-39) / (x^2+2x-4)$

$= (5 x^{2} - 12 x + 37) - \frac{122 x + 187}{x^{2} + 2 x - 4}$ $= (5x^2-12x+37) - (122x+187)/(x^2+2x-4)$

How do you long divide 5x4−2x3−7x2−39x2+2x−4(5x^4-2x^3- 7x^2 -39) / (x^2 + 2x- 4)?