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.

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