How do you divide #(x^4 + 5x^2 + 12x + 3)/( 6x+8)#?

1 Answer
Jun 9, 2016

#=> x^3/6-(2x^2)/9+(61x)/54+40/81-77/(486x+648)#

Explanation:

#" "x^4+0x^3+5x^2+12x+3#
#color(magenta)(1/6 x^3)(6x+8) -> " "color(white)(a)ul(x^4+4/3 x^3" ")" "larr "Subtract"#
#" "0 - 4/3x^3 +5x^2+12x+3#
#color(magenta)(-2/9x^2)(6x+8) ->" "color(white)(a)ul(-4/3x^3-16/9x^2" ")" "larr "Subtract"#
#" "0 + 61/9x^2+12x+3#
#color(magenta)(61/54x)(6x+8) ->" "color(white)(a)ul(61/9x^2+244/27x)" "larr "Subtract"#
#" "0+80/27x+3#
#color(magenta)(40/81)(6x+8) =>" "color(white)(a)ul(80/27x+320/81)" Subtract"#
#" "0-77/81#

#=> x^3/6-(2x^2)/9+(61x)/54+40/81-77/(81(6x+8)#

#=> x^3/6-(2x^2)/9+(61x)/54+40/81-77/(486x+648)#