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"#
