How do you divide #(x^4-3x^3+5x-6)div (x+2)#?

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Answer 1 · 2016-05-14T16:37:28

#"Quotient"=(x^3-5x^2+10x-15)#
#"Remainder"=24#

#(x^4-3x^3+5x-6)div (x+2)#

#=(x^4-3x^3+5x-6)/ (x+2)#

#=(x^3(x+2)-2x^3-3x^3+5x-6)/ (x+2)#

#=(x^3(x+2)-5x^3+5x-6)/ (x+2)#

#=(x^3(x+2)-5x^2(x+2)+10x^2+5x-6)/ (x+2)#

#=(x^3(x+2)-5x^2(x+2)+10x(x+2)-20x+5x-6)/ (x+2)#

#=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15x-6)/ (x+2)#

#=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15(x+2)+30-6)/ (x+2)#

#=(x^3(x+2)-5x^2(x+2)+10x(x+2)-15(x+2)+24)/ (x+2)#

#=(cancel((x+2))(x^3-5x^2+10x-15))/cancel((x+2))+24/ (x+2)#

#=(x^3-5x^2+10x-15)+24/ (x+2)#

Result
#"Quotient"=(x^3-5x^2+10x-15)#
#"Remainder"=24#