How do you long divide #(2x^2+10x+12)# by #x+3#?

1 Answer
Oct 29, 2015

#2x+4#

Explanation:

#color(green)("Trying to build the old school format for long division produces an unexpected format")#. I do not think this web system was built to cope with what I am trying to achieve. I will do the best I can! Long division is all about dividing sections of the original equation by appropriate section of the divisor and then dealing with the remainder.

The first part will be from dividing only the #x# parts so we have #2x^2 divide x# of #x+3#.

#color(red)("Division result")##" "-> color(blue)(2x) + color(red)(4)#

#color(brown)("Step 1.")#
#2x^2 divide x#
giving #color(blue)(2x)# as part answer. This is the 1st part of 'division result'
shown above.

Remainder #(2x^2+10x+12)-2x(x+3)=4x+12#

#color(brown)("Step 2.")#
Repeat the process on the remainder and again only divide
the #x# parts

#4x divide x = (+4)#
giving #color(red)(4)# as part answer. This is the 2nd part of 'division result'
shown above.

Remainder: #(4x+12) - 4(x+3) = 0

So #(x+3)# divides exactly into #(2x^2 +10x+12)#

If this helps please click on the thumbs up and let me know. Thank you!