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

Follow the colours.

#(n^3+7n^2 +14n+3)div(n+2) = ?????????#

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

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

#color(white)(xxxxxxxxxxx)n^3 +7n^2 +14n +3#

#color(white)(xxxxxxxx) rArr color(magenta)(1" +7 +14 +3") #

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. # " " (n+2) = 0 rArr n = color(orange)(-2) " this goes outside"#

#color(white)(xxxxx) | color(brown)(1)" "+7" "+14 " "+3 color(magenta)(" step 1 top row")#

#color(white)(x.x)color(orange)(-2) ""| darr " "color(red)(-2) " "color(blue)(-10) " "color(olive)(-8)#

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

#color(white)(xxxxxxx) color(brown)(1) " "color(blue)(+5) " "color(olive)(+4)" "color(teal)(-5) larr " remainder!"#

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

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

**Step 3** : Begin the division:

#"Bring down the " color(brown)( 1 ) " to below the line"#

#"multiply " color(orange)(-2) xx color(brown)(1) = color(red)(-2)#

#"Add " 7+color(red)(-2) = color(blue)(+5)#

#"multiply " color(orange)(-2) xx color(blue)(+5) = color(blue)(-10)#

#"Add " 14 color(blue)( -10) = color(olive)(+4)#

#"multiply " color(orange)(-2) xxcolor(olive)(4) = color(olive)(-8)#

#"Add " 3 +color(olive)(-8) = color(teal)(-5)#

That's it Folks!

We have now found the numerical coefficients of the terms in the quotient (answer)

We divided an expression with #n^3# by an expression with #n#,

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

The last value is the remainder. In this case it is #color(teal)(-5)#

This means that #(n+2)# is not a factor of #n^3 +7n^2+14n+3#

#(n^3 +7n^2+14n+3) div(n+2) = n^2 +5n +4 " rem -5"#