How do you divide: #5a^2+6a-9# into #25a^4#?

1 Answer
George C.
Jun 5, 2015

Use synthetic division. The process is somewhat like long division.

First choose a multiplier of #5a^2+6a-9# which will match the most significant part of #25a^4#.

That first multiplier is #color(red)(5a^2)#.

#5a^2 (5a^2+6a-9) = 25a^4+30a^3-45a^2#

Subtract this from the original polynomial to get a remainder...

#25a^4 - (25a^4+30a^3-45a^2) = -30a^3+45a^2#

Next choose a second multiplier to match the leading term of this remainder...

The second multiplier is #color(red)(-6a)#

#-6a(5a^2+6a-9) = -30a^3+36a^2+54a#

Subtract this from the remainder to get a new remainder...

#(-30a^3+45a^2) - (-30a^3+36a^2+54a)#

#=9a^2-54a#

Next choose a third multiplier to match the leading term of this remainder...

The third multiplier is #color(red)(9/5)#

#9/5(5a^2+6a-9) = 9a^2+54/5a-81/5#

Subtract this from the previous remainder to get a new remainder...

#(9a^2-54a) - (9a^2+54/5a-81/5)#

#=-(54+54/5)a+81/5#

#=-(270+54)/5a+81/5#

#=(81-324a)/5#

#=81/5(1-4a)#

Adding all the multipliers we found together, we have:

#(25a^4) / (5a^2+6a-9)#

#= (5a^2-6a+9/5)+81/5*(1-4a)/(5a^2+6a-9)#

I think this is where you are expected to stop.

Like long division, you could carry on to find terms in #a^(-1)#, #a^(-2)#, etc., like the decimal places of a long division of numbers, but the result would probably not be very useful from an algebra perspective.

Related questions
See all questions in Division of Polynomials
Impact of this question
4766 views around the world
You can reuse this answer
Creative Commons License