Question

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

Answer 1

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.