How do you divide #(12x^3+2+11x+20x^2) / (2x+1)#?

Question

3646 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-23T12:03:44

#6x^2+7x+2#

Rearrange the top expression to standard format

#(12x^3+20x^2+11x+2)/(2x+1)#

I prefer to set the question as long division. How many #2x#s can you take from #12x^3? 12/2 = 6, x^3/x = x^2.# So, #6x^2#

Multiply the divisor #2x+1# by #6x^2# and subtract.
#(2x+1)xx(6x^2)=(12x^3+6x^2)#
#(12x^3+20x^2+11x+2)-(12x^3+6x^2)=(14x^2+11x+2)#

How many #2x#s can you take from #14x^2#? .....#7x#

Multiply the divisor #2x+1# by #7x# and subtract.
#(14x^2+11x+2)-(14x^2+7x)=(4x+2)#

How many #2x#s can you take from #4x#?.....2

Multiply the divisor #2x+1# by #2# and subtract.
#(4x+2)-(4x+2) = 0#