Question 1204260
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Leading coefficient = 12, which is always attached to the term with the largest exponent
Constant term = -8


Positive factors of the constant term: 1, 2, 4, 8
Positive factors of the leading coefficient: 1, 2, 3, 4, 6, 12


Each possible rational root will be of the form p/q
p = factor of the constant term
q = factor of the leading term


The way I like to do this kind of problem is to arrange the factors mentioned in a table like this
<table border = "1" cellpadding = "5"><tr><td></td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>1</td><td></td><td></td><td></td><td></td></tr><tr><td>2</td><td></td><td></td><td></td><td></td></tr><tr><td>3</td><td></td><td></td><td></td><td></td></tr><tr><td>4</td><td></td><td></td><td></td><td></td></tr><tr><td>6</td><td></td><td></td><td></td><td></td></tr><tr><td>12</td><td></td><td></td><td></td><td></td></tr></table>


Then to fill out the first row, we divide the items in the upper row header (1,2,4 and 8) by the value 1.
The values won't change.
<table border = "1" cellpadding = "5"><tr><td></td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>1</td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>2</td><td></td><td></td><td></td><td></td></tr><tr><td>3</td><td></td><td></td><td></td><td></td></tr><tr><td>4</td><td></td><td></td><td></td><td></td></tr><tr><td>6</td><td></td><td></td><td></td><td></td></tr><tr><td>12</td><td></td><td></td><td></td><td></td></tr></table>
The second row is a bit more interesting. We divide the items in the upper row (1,2,4,8) by the value 2. Each value in the second row is fully reduced. Eg: 2/2 reduces to 1.
<table border = "1" cellpadding = "5"><tr><td></td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>1</td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>2</td><td>1/2</td><td>1</td><td>2</td><td>4</td></tr><tr><td>3</td><td></td><td></td><td></td><td></td></tr><tr><td>4</td><td></td><td></td><td></td><td></td></tr><tr><td>6</td><td></td><td></td><td></td><td></td></tr><tr><td>12</td><td></td><td></td><td></td><td></td></tr></table>
This process is carried out to have this completed table.
<table border = "1" cellpadding = "5"><tr><td></td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>1</td><td>1</td><td>2</td><td>4</td><td>8</td></tr><tr><td>2</td><td>1/2</td><td>1</td><td>2</td><td>4</td></tr><tr><td>3</td><td>1/3</td><td>2/3</td><td>4/3</td><td>8/3</td></tr><tr><td>4</td><td>1/4</td><td>1/2</td><td>1</td><td>2</td></tr><tr><td>6</td><td>1/6</td><td>1/3</td><td>2/3</td><td>4/3</td></tr><tr><td>12</td><td>1/12</td><td>1/6</td><td>1/3</td><td>2/3</td></tr></table>
At first glance it appears there are 4*6 = 24 possible positive rational roots. 


However, there are a few repeated items. For instance, "2/3" shows up 3 times.
After removing duplicates, we have these possible positive rational roots: 
1, 2, 4, 8, 1/2, 1/3, 2/3, 4/3, 8/3, 1/4, 1/6, 1/12


List the plus and minus version of each root to get the full set of possible rational roots.
Of course, U(x) will only have up to 5 rational roots since this is the degree of the polynomial (it's the largest exponent).


Answer: 
<font color=red size=4>±1, ±2, ±4, ±8, ±1/2, ±1/3, ±2/3, ±4/3, ±8/3, ±1/4, ±1/6, ±1/12</font>


Feel free to find other ways to arrange the values to what makes the most sense for you.
The way I've arranged them is to put the whole numbers first, then fractions afterward. 
The denominators are grouped together. Within the subgroup that has denominator 3, the numerators increase from left to right.


A lengthier way to write the list would be to say:
<font color=red>1, -1,  2, -2,  4, -4,  8, -8,  1/2, -1/2,  1/3, -1/3,  2/3, -2/3,  4/3, -4/3,  8/3, -8/3,  1/4, -1/4,  1/6, -1/6,  1/12, -1/12</font>
The first method is more efficient since we don't have to write the same value twice more or less.
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