How do you divide #(-x^5-4x^3+x-12)/(x^2-x+3)#?

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Answer 1 · 2018-07-10T06:53:37

The remainder is #=(8x-15)# and the quotient is #=(-x^3-x^2-2x+1)#

Perform a long division

#color(white)(aa)##-x^5+0x^4-4x^3+0x^2+x-12##color(white)(aa)##|##x^2-x+3#

#color(white)(aa)##-x^5+x^4-3x^3##color(white)(aaaaaaaaaaaaaaaa)##|##-x^3-x^2-2x+1#

#color(white)(aaaa)##0-1x^4-1x^3+0x^2#

#color(white)(aaaaaa)##-1x^4+1x^3-3x^2#

#color(white)(aaaaaaaaa)##0-2x^3+3x^2+x#

#color(white)(aaaaaaaaaaa)##-2x^3+2x^2-6x#

#color(white)(aaaaaaaaaaaaaa)##0+x^2+7x-12#

#color(white)(aaaaaaaaaaaaaaaa)##+x^2-1x+3#

#color(white)(aaaaaaaaaaaaaaaaa)##-0+8x-15#

Therefore,

#(-x^5+0x^4-4x^3+0x^2+x-12)/(x^2-x+3)#

#=(-x^3-x^2-2x+1)+(8x-15)/(x^2-x+3)#

The remainder is #=(8x-15)# and the quotient is #=(-x^3-x^2-2x+1)#