The method is easy, but the process is a bit difficult to explain.
Follow the colours.
#" "(x^2+2x+15)div(x-3)#
#" (dividend) " div " (divisor)"#
#color(magenta)("step 1:")# The dividend must be in descending powers of x.
#color(white)(xxxxxxxxxxxxxxxxxxxxxxxx)x^2 " "+2x" " +15#
Only use the numerical coefficients # rArr 1" "+2" "+15 "#
(If there are any missing, leave a space or fill in a zero).
#color(orange)("Step 2")#: Make the divisor = 0. # " "(x-3) = 0 rArr x = color(orange)(3) " this goes outside"#
#color(white)(xxxxxxxxx) | color(brown)(1)" "+2" "+15" "color(magenta)("step 1")#
#color(white)(xxxxxx)color(orange)(3) " "| darr " "color(red)(3) " "color(blue)(15)#
#color(white)(xxxxxxxxxx) ul(" ")#
#color(white)(xxxxxxxxxxx) color(brown)(1) " "color(blue)(5) " "color(teal)(30) larr "remainder!"#
#color(white)(xxxx.. xxxxx)uarr " "uarr#
#color(white)(xxxxxxxxxxx) x " "x^0#
Step 3: Begin the division:
#"Bring down the " color(brown)( 1 ) " to below the line"#
#"multiply " color(orange)(3) xx color(brown)(1) = color(red)(3)#
#"Add " 2+color(red)(3) = color(blue)(5)#
#"multiply " color(orange)(3) xx color(blue)(5) = color(blue)(15)#
#"Add" 15+color(blue)(15) = color(teal)(30)#
That's it Folks!
We have now found the numerical coefficients of the terms in the quotient (answer)
We divided an expression with #x^2# by an expression with #x#,
so the first term will be #x^2/x = x#
The last value is the remainder. In this case it is #color(teal)(30)#
This means that #x-3# is not a factor of #x^2 +2 +15#
#(x^2 +2x +15) div (x-3) = color(brown)(1)x +color(blue)(5), "rem " color(teal)(30)#
