The method is very easy, but the process is a bit difficult to explain.
Follow the colours.
#(14y+8y^2+y^3+12)div(6+y) = ?????????#
#" (dividend) " div " (divisor)" = ("quotient")#
#color(magenta)("step 1:")# The dividend must be in descending powers of r.
#color(white)(xxxxxxxxxxx)y^3 +8y^2 +14y +12#
#color(white)(xxxxxxxx) rArr color(magenta)(1" +8 +14 +12") #
Only use the numerical coefficients in the #color(magenta)("top row")#.
(If there are any missing, leave a space or fill in a zero).
#color(orange)("Step 2")#: Make the divisor = 0. # " " (6+y) = 0 rArr y = color(orange)(-6) " this goes outside"#
#color(white)(xxxxx) | color(brown)(1)" "+8" "+14 " "+12 color(magenta)(" step 1 top row")#
#color(white)(x.x)color(orange)(-6) ""| darr " "color(red)(-6) " "color(blue)(-12) " "color(olive)(-12)#
#color(white)(xxxxxx) ul(" ")#
#color(white)(xxxxxxx) color(brown)(1) " "color(blue)(+2) " "color(olive)(+2)" "color(teal)(0) larr " no remainder!"#
#color(white)(xxxx.xx)uarr " "uarr " "uarr#
#color(white)(xxxxxxx) y^2 " "y^1 " "y^0#
Step 3 : Begin the division:
#"Bring down the " color(brown)( 1 ) " to below the line"#
#"multiply " color(orange)(-6) xx color(brown)(1) = color(red)(-6)#
#"Add " 8+color(red)(-6) = color(blue)(+2)#
#"multiply " color(orange)(-6) xx color(blue)(+2) = color(blue)(-12)#
#"Add " 14 color(blue)( -12) = color(olive)(+2)#
#"multiply " color(orange)(-6) xxcolor(olive)(2) = color(olive)(-12)#
#"Add " 12 +color(olive)(-12) = color(teal)(0)#
That's it Folks!
We have now found the numerical coefficients of the terms in the quotient (answer)
We divided an expression with #y^3# by an expression with #y#,
so the first term will be #y^3/y = y^2#
The last value is the remainder. In this case it is #color(teal)(0)#
This means that #(y+6)# is a factor of #y^3 +8y^2+14y+12#
#(y^3 +8y^2+14y+12) div(y+6) = y^2 +2y +2 " rem 0"#
