Use synthetic division. The process is somewhat like long division.
First choose a multiplier of 5a^2+6a-95a2+6a−9 which will match the most significant part of 25a^425a4.
That first multiplier is color(red)(5a^2)5a2.
5a^2 (5a^2+6a-9) = 25a^4+30a^3-45a^25a2(5a2+6a−9)=25a4+30a3−45a2
Subtract this from the original polynomial to get a remainder...
25a^4 - (25a^4+30a^3-45a^2) = -30a^3+45a^225a4−(25a4+30a3−45a2)=−30a3+45a2
Next choose a second multiplier to match the leading term of this remainder...
The second multiplier is color(red)(-6a)−6a
-6a(5a^2+6a-9) = -30a^3+36a^2+54a−6a(5a2+6a−9)=−30a3+36a2+54a
Subtract this from the remainder to get a new remainder...
(-30a^3+45a^2) - (-30a^3+36a^2+54a)(−30a3+45a2)−(−30a3+36a2+54a)
=9a^2-54a=9a2−54a
Next choose a third multiplier to match the leading term of this remainder...
The third multiplier is color(red)(9/5)95
9/5(5a^2+6a-9) = 9a^2+54/5a-81/595(5a2+6a−9)=9a2+545a−815
Subtract this from the previous remainder to get a new remainder...
(9a^2-54a) - (9a^2+54/5a-81/5)(9a2−54a)−(9a2+545a−815)
=-(54+54/5)a+81/5=−(54+545)a+815
=-(270+54)/5a+81/5=−270+545a+815
=(81-324a)/5=81−324a5
=81/5(1-4a)=815(1−4a)
Adding all the multipliers we found together, we have:
(25a^4) / (5a^2+6a-9)25a45a2+6a−9
= (5a^2-6a+9/5)+81/5*(1-4a)/(5a^2+6a-9)=(5a2−6a+95)+815⋅1−4a5a2+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−1, a^(-2)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.
