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) = ????????? (8r3−55r2+44r−12)÷(r−6)=?????????
" (dividend) " div " (divisor)" = ("quotient") (dividend) ÷ (divisor)=(quotient)
color(magenta)("step 1:")step 1: The dividend must be in descending powers of r.
color(white)(xxxxxxxxxx)8r^3 " -55r^2 +44r -12×××××8r3−55r2+44r−12
color(white)(xxxxxxxx) rArr 8" -55 +44 -12"××××⇒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")Step 2: Make the divisor = 0. " " (r-6) = 0 rArr r = color(orange)(6) " this goes outside" (r−6)=0⇒r=6 this goes outside
color(white)(xxxxx) | color(brown)(8)" "-55" "+44 " "-12 color(magenta)(" step 1")××x∣8 −55 +44 −12 step 1
color(white)(xx)color(orange)(6) " "| darr " "color(red)(48) " "color(blue)(-42) " "color(olive)(12)×6 ∣⏐↓ 48 −42 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"
