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)#