How do you divide #(4x^2-3x+9)/(x+4)#?

1 Answer
Jun 14, 2016

A different presentation style

#(4x^2-3x+9)/(x+4)" "=" "4x-3+5/(x+4)#

Explanation:

#" "color(blue)(4x^2-3x+9)#
#color(magenta)(4x)(x+4)->" "ul(4x^2+16x)" "larr" Subtract"#
#" " 0-19x#
#ul(color(magenta)(-19)(x+4)->" "-19x-76" "larr" Subtract"#
#"Remainder "->" "0-76#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(4x^2-3x+9)/(x+4)" "=" "4x-19-76/(x+4)#