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

1 Answer
May 14, 2016

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

Explanation:

(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