How do you divide #(6x^2+x-7)div(2x+3)# using long division?

1 Answer
Aug 24, 2016

#" Dividend" div "divisor" = "quotient"#
#(6x^2+x-7)div(2x+3) = (3x-4) " remainder 5"#

Explanation:

#" Dividend" div "divisor" = "quotient"#
#(6x^2+x-7)div(2x+3) = (3x-4) " remainder 5"#

The dividend must be in descending powers of x.

In long division: repeat the pattern:
Divide #color(magenta)(2x)# into the term with highest index available
Multiply by both terms at the side
Subtract (change the signs)
Bring down the next term

#color(white)(xxxx)ulcolor(white)(xxxxx)ul(color(red)(3x)color(blue)(-4))#
#color(magenta)(2x)+3 )6x^2+x-7 " "6x^2 divcolor(magenta)(2x) = color(red)(3x)#
#color(white)(xxxx)ulcolor(red)(-6x -9x)" "darr " multiply and subtract"#
#color(white)(xxxxxxx)-8x -7" "-8x divcolor(magenta)(2x)= color(blue)(-4)#
#color(white)(xxxxxxxxx)ulcolor(blue)(8x+12)" multiply and subtract"#
#color(white)(xxxxxxxxxxxxx)5" remainder"#