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

Jun 5, 2015

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

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

That first multiplier is $\textcolor{red}{5 {a}^{2}}$.

$5 {a}^{2} \left(5 {a}^{2} + 6 a - 9\right) = 25 {a}^{4} + 30 {a}^{3} - 45 {a}^{2}$

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

$25 {a}^{4} - \left(25 {a}^{4} + 30 {a}^{3} - 45 {a}^{2}\right) = - 30 {a}^{3} + 45 {a}^{2}$

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

The second multiplier is $\textcolor{red}{- 6 a}$

$- 6 a \left(5 {a}^{2} + 6 a - 9\right) = - 30 {a}^{3} + 36 {a}^{2} + 54 a$

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

$\left(- 30 {a}^{3} + 45 {a}^{2}\right) - \left(- 30 {a}^{3} + 36 {a}^{2} + 54 a\right)$

$= 9 {a}^{2} - 54 a$

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

The third multiplier is $\textcolor{red}{\frac{9}{5}}$

$\frac{9}{5} \left(5 {a}^{2} + 6 a - 9\right) = 9 {a}^{2} + \frac{54}{5} a - \frac{81}{5}$

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

$\left(9 {a}^{2} - 54 a\right) - \left(9 {a}^{2} + \frac{54}{5} a - \frac{81}{5}\right)$

$= - \left(54 + \frac{54}{5}\right) a + \frac{81}{5}$

$= - \frac{270 + 54}{5} a + \frac{81}{5}$

$= \frac{81 - 324 a}{5}$

$= \frac{81}{5} \left(1 - 4 a\right)$

Adding all the multipliers we found together, we have:

$\frac{25 {a}^{4}}{5 {a}^{2} + 6 a - 9}$

$= \left(5 {a}^{2} - 6 a + \frac{9}{5}\right) + \frac{81}{5} \cdot \frac{1 - 4 a}{5 {a}^{2} + 6 a - 9}$

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.