How do you long divide #(3y^3 + 4y^2 - 5y - 6)/( y + 2)#?

1 Answer
Aug 25, 2016

#" Dividend" div "divisor" = "quotient"#
#(3y^3 +4y^2-5y-6)div(2y+2) = (3y^2 -2y-1) " remainder -4"#

Explanation:

#" Dividend" div "divisor" = "quotient"#
#(3y^3 +4y^2-5y-6)div(2y+2) = (3y^2 -2y-1) " remainder -4"#

The dividend must be in descending powers of y.

In long division: repeat the following sequence:

Divide #color(magenta)(y)# into the term with highest index available
Multiply by both terms at the side #(y+2)#
Subtract (change the signs)
Bring down the next term

#color(white)(xxxx)ulcolor(white)(xxxxx)ul(color(red)(3y^2)color(blue)(-2y)color(teal)(-1) " " rem -4#
#color(magenta)(y)+2)3y^3 +4y^2-5y-6" "3y^3 divcolor(magenta)(y) = color(red)(3y^2)#
#color(white)(xxxx)ulcolor(red)(3y^3+6y^2)" multiply and subtract"#
#color(white)(xxxxxxx)-2y^2 -5y" "-2y^2 divcolor(magenta)(y)= color(blue)(-2y)#
#color(white)(xxxxx.xx)ulcolor(blue)(-2y^2-4y)" multiply and subtract"#
#color(white)(xxxx.xxxxxxx)-y-6" "-y div color(magenta)(y) = color(teal)(-1) #
#color(white)(xxxx..xxxxxxx)ul(-y-2)" multiply and subtract"#
#color(white)(xxxx..xxxxxxxxx)-4 " remainder"#