Question 813301
{{{3x^3+x^2-15x-5}}}
The possible rational roots of a polynomial are all the ratios, positive and negative, that can be formed using a factor of the constant term in the numerator and a factor of the leading coefficient in the denominator.<br>
Your constant term is 5. (Actually it is -5 but since we will be using all positive and negative ratios it doesn't make any difference if we use 5 or -5.) The factors of 5 are just 1 and 5.<br>
Your leading coefficient is 3 whose factors are just 1 and 3.<br>
Forming the ratios (as described above) we get the following list of possible rational roots:
<u>+</u>1/1, <u>+</u>5/1, <u>+</u>1/3 and <u>+</u>5/3.
The first two simplify:
<u>+</u>1, <u>+</u>5, <u>+</u>1/3 and <u>+</u>5/3.<br>
Now we see which of these, if any, are actual roots. We'll use synthetic division to check. Checking 1:
<pre>
1  |   3   1   -15   -5
----       3     4  -11
      ------------------
       3   4   -11  -16
</pre>The remainder is the -16 in the lower right corner. Since it is not zero 1 is not a root. Checking x = -1:
<pre>
-1 |   3   1   -15   -5
----      -3     2   13
      ------------------
       3  -2   -13    8
</pre>The remainder is the 8. So -1 is not a root, either. But since this remainder is positive and the one we got for 1 was negative, there is a root somewhere in between. Since 8 is closer to zero than -16 I'm guessing that the root is closer to -1 than to 1. So I will try -1/3 next:
<pre>
-1/3 |   3   1   -15   -5
------      -1     0    5
        ------------------
         3   0   -15    0
</pre>And we have a winner! -1/3 is a root (and 3x+1 is a factor). Not only that, but the rest of the bottom row tells us the other factor. The "3  0  -15" translates into {{{3x^2+0x-15}}} or simply {{{3x^2-15}}}. Since this is a quadratic we can find the remaining roots without having to try more rational roots. All we have to do is solve:
{{{3x^2-15=0}}}
Adding 15:
{{{3x^2=15}}}
Dividing by 3: 
{{{x^2 = 5}}}
Square root of each side:
{{{x = 0+-sqrt(5)}}}
(Note: The zero is there only because algebra.com's formula drawing software will not let me use the "plus or minus" symbol without a number in front.)<br>
So the three roots are -1/3 (rational), {{{sqrt(5)}}} (irrational) and {{{-sqrt(5)}}} (irrational)