The method is easy, but the process is a bit difficult to explain.
Follow the colours.
" "(x^2+2x+15)div(x-3) (x2+2x+15)÷(x−3)
" (dividend) " div " (divisor)" (dividend) ÷ (divisor)
color(magenta)("step 1:")step 1: The dividend must be in descending powers of x.
color(white)(xxxxxxxxxxxxxxxxxxxxxxxx)x^2 " "+2x" " +15××××××××××××x2 +2x +15
Only use the numerical coefficients rArr 1" "+2" "+15 "⇒1 +2 +15
(If there are any missing, leave a space or fill in a zero).
color(orange)("Step 2")Step 2: Make the divisor = 0. " "(x-3) = 0 rArr x = color(orange)(3) " this goes outside" (x−3)=0⇒x=3 this goes outside
color(white)(xxxxxxxxx) | color(brown)(1)" "+2" "+15" "color(magenta)("step 1")××××x∣1 +2 +15 step 1
color(white)(xxxxxx)color(orange)(3) " "| darr " "color(red)(3) " "color(blue)(15)×××3 ∣⏐↓ 3 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)
