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"
