How do you simplify $\frac{6 x^{2} - x + 6}{2 x - 1}$ $(6x^2-x+6 )/( 2x-1 )$ ?

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Answer 1 · 2015-05-17T06:57:51

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 $3 x (2 x - 1) = 6 x^{2} - 3 x$ $3x(2x - 1) = 6x^2-3x$ , which will allow us to separate the $x^{2}$ $x^2$ term:

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

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

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

$= 2 x + 6$ $=2x + 6$

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

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

$1 \times (2 x - 1) = 2 x - 1$ $1xx(2x-1) = 2x-1$

Then

$(2 x + 6) - (2 x - 1) = 2 x + 6 - 2 x + 1 = 7$ $(2x + 6) - (2x - 1) = 2x + 6 - 2x + 1 = 7$

This gives $7$ $7$ 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 $(3 x + 1)$ $(3x+1)$ .

So to summarize where we have got to so far:

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

Therefore

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

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

How do you simplify 6x2−x+62x−1(6x^2-x+6 )/( 2x-1 )?