The method is very easy, but the process is a bit difficult to explain.

Follow the colours.

#(8r^3-55r^2+44r-12)div(r-6) = ????????? #

#" (dividend) " div " (divisor)" = ("quotient")#

#color(magenta)("step 1:")# The dividend must be in descending powers of r.

#color(white)(xxxxxxxxxx)8r^3 " -55r^2 +44r -12#

#color(white)(xxxxxxxx) rArr 8" -55 +44 -12"#

Only use the numerical coefficients in the top row.

(If there are any missing, leave a space or fill in a zero).

#color(orange)("Step 2")#: Make the divisor = 0. # " " (r-6) = 0 rArr r = color(orange)(6) " this goes outside"#

#color(white)(xxxxx) | color(brown)(8)" "-55" "+44 " "-12 color(magenta)(" step 1")#

#color(white)(xx)color(orange)(6) " "| darr " "color(red)(48) " "color(blue)(-42) " "color(olive)(12)#

#color(white)(xxxxxx) ul(" ")#

#color(white)(xxxxxxx) color(brown)(8) " "color(blue)(-7) " "color(olive)(+2)" "color(teal)(0) larr " no remainder!"#

#color(white)(xxxx.xx)uarr " "uarr " "uarr#

#color(white)(xxxxxxx) r^2 " "r^1 " "r^0#

**Step 3** : Begin the division:

#"Bring down the " color(brown)( 8 ) " to below the line"#

#"multiply " color(orange)(6) xx color(brown)(8) = color(red)(48)#

#"Add " -55+color(red)(48) = color(blue)(-7)#

#"multiply " color(orange)(6) xx color(blue)(-7) = color(blue)(-42)#

#"Add " 44 color(blue)( -42) = 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 #r^3# by an expression with #r#,

so the first term will be #r^3/r = r^2#

The last value is the remainder. In this case it is #color(teal)(0)#

This means that #(r-6)# is a factor of #8r^3-55r^2+44r-12#

#(8r^3-55r^2+44r-12) div(r-6) = 8r^2-7r+2 " rem 0"#