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

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Answer 1 · 2015-06-06T23:44:29

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

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

Subtract this from the original numerator to get a remainder:

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

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

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

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

Subtract this from the remainder to get a new remainder:

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

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

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

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

Subtract this from the remainder to get a new remainder:

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

$=-122x-187$

Add the multiplier we have found together to get:

$(5x^4-2x^3-7x^2-39) / (x^2+2x-4)$

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

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