# How do you simplify (6x^2-x+6 )/( 2x-1 )?

May 17, 2015

Use synthetic division:

I will write it up a little verbosely to make the process as clear as possible. Essentially it's like doing long division.

First notice that $3 x \left(2 x - 1\right) = 6 {x}^{2} - 3 x$, which will allow us to separate the ${x}^{2}$ term:

$\left(6 {x}^{2} - x + 6\right) - 3 x \left(2 x - 1\right)$

$= \left(6 {x}^{2} - x + 6\right) - \left(6 {x}^{2} - 3 x\right)$

$= 6 {x}^{2} - x + 6 - 6 {x}^{2} + 3 x$

$= 2 x + 6$

So $3 x$ makes a good multiplier, with remainder #(2x+6).

To match the leading term $2 x$ in the remainder, the next multiplier we want is $1$:

$1 \times \left(2 x - 1\right) = 2 x - 1$

Then

$\left(2 x + 6\right) - \left(2 x - 1\right) = 2 x + 6 - 2 x + 1 = 7$

This gives $7$ as our final remainder. If the remainder was zero then we would have a simple factorization.

Add the multipliers that we have found together to get the term $\left(3 x + 1\right)$.

So to summarize where we have got to so far:

$\left(6 {x}^{2} - x + 6\right) = \left(3 x + 1\right) \left(2 x - 1\right) + 7$

Therefore

$\frac{6 {x}^{2} - x + 6}{2 x - 1} = \left(3 x + 1\right) + \frac{7}{2 x - 1}$

$= 3 x + 1 + \frac{7}{2 x - 1}$