Question 806049
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Polynomial long division is a valid method as shown by tutor @josgarithmetic. 


Synthetic division is the more efficient route (in my opinion), which is what I'll use below.


The denominator x-2 has the test root x = 2
We write 2 in the upper left corner.


Think of x^3-6 as 1x^3+0x^2+0x+(-6) so we can see the coefficients much more clearly. Those coefficients are: 1, 0, 0, -6 which are laid out across the top row (ignoring the previous "2" mentioned earlier)
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr></table>


We'll drop the 1 down like so
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td>1</td><td></td><td></td><td></td></tr></table>
Then multiply the test root (2) with that value we dropped. The result 2*1 = 2 is placed just under the 0 in the next column
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td>2</td><td></td><td></td></tr><tr><td></td><td>1</td><td></td><td></td><td></td></tr></table>


Next we add the stuff along the 2nd column: 0+2 = 2
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td>2</td><td></td><td></td></tr><tr><td></td><td>1</td><td>2</td><td></td><td></td></tr></table>


Multiply the test root (2) with this new value in the bottom row: 2*2 = 4
Write 4 just under the other 0
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td>2</td><td>4</td><td></td></tr><tr><td></td><td>1</td><td>2</td><td></td><td></td></tr></table>



Keep this process going until the table is filled out. 
This is what you should get:
<table border = "1" cellpadding = "5"><tr><td>2</td><td>1</td><td>0</td><td>0</td><td>-6</td></tr><tr><td></td><td></td><td>2</td><td>4</td><td>8</td></tr><tr><td></td><td>1</td><td>2</td><td>4</td><td>2</td></tr></table>


If you are confused with any step of the synthetic division process, then please let me know. I recommend searching out videos of similar examples. Just so you get a feel of what's going on. It's a bit tricky to describe the movement patterns with text only.


The last number in the bottom row is the remainder.
The other numbers in the bottom row are the coefficients of the quotient.
The values 1, 2, 4 are the coefficients to 1x^2+2x+4 aka x^2+2x+4


This is how we end up with 
quotient = x^2+2x+4
remainder = 2


The remainder 2 is placed over the denominator (x-2)
Which means we can write
{{{(x^3-6)/(x-2) = x^2+2x+4 + 2/(x-2)}}}
This is an identity which is a true equation for almost all real numbers x as long as {{{x <> 2}}}


Side note: Because the remainder is not zero, it means (x-2) is not a factor of x^3-6.
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