<PRE><B>Question 813325</B>
{{{10x^4+x^3+7x^2+x-3}}}
To find the roots we need to factor the expression. With a GCF of 1 and these exponents and this many terms, about the only factoring method that will work (to start) is where you determine the possible rational roots and try them out (hoping to find one or more).&lt;br&gt;
The possible rational roots of a polynomial will be all the ratios, positive and negative, which can be formed using a factor of the constant term in the numerator and a factor of the leading coefficient in the denominator. The constant term in this polynomial is 3. (Actually it is -3 but since we&#39;re going to include all positive and negative ratios, it does not make any difference whether we use 3 or -3 for this.) The factors of 3 are just 1 and 3. The leading coefficient is 10. The factors of 10 are 1, 2, 5 and 10. This makes the list of possible rational roots:
&lt;u&gt;+&lt;/u&gt;1/1, &lt;u&gt;+&lt;/u&gt;3/1, &lt;u&gt;+&lt;/u&gt;1/2, &lt;u&gt;+&lt;/u&gt;3/2, &lt;u&gt;+&lt;/u&gt;1/5, &lt;u&gt;+&lt;/u&gt;3/5, &lt;u&gt;+&lt;/u&gt;1/10 and &lt;u&gt;+&lt;/u&gt;3/10
The first two simplify leaving us with:
&lt;u&gt;+&lt;/u&gt;1, &lt;u&gt;+&lt;/u&gt;3, &lt;u&gt;+&lt;/u&gt;1/2, &lt;u&gt;+&lt;/u&gt;3/2, &lt;u&gt;+&lt;/u&gt;1/5, &lt;u&gt;+&lt;/u&gt;3/5, &lt;u&gt;+&lt;/u&gt;1/10 and &lt;u&gt;+&lt;/u&gt;3/10&lt;br&gt;
Testing these possible roots to see if they are actual roots is often done using synthetic division. I&#39;ll start with 1:
&lt;pre&gt;
1  |   10   1   7   1   -3
----       10  11  18   19
      ---------------------
       10  11  18  19   16
&lt;/pre&gt;The number in the lower right corner, the 16, is the remainder. It is not zero so 1 is not a root. And since 16 is not very close to zero and since this polynomial&#39;s end behavior tells me that we may have already reached the end behavior. So I am going to skip testing any higher x&#39;s for now. I may come back to them later. Also, since the polynomial is equal to -3 when x is zero, I know that between x = 0 (where the y is negative) and x = 1 (where the y is positive), I know that there is a root between 0 and 1. So I will try some fractions in that interval. We&#39;ll try 1/2 first:
&lt;pre&gt;
1/2 |   10   1   7   1   -3
-----        5   3   5    3
       ---------------------
        10   6  10   6    0
&lt;/pre&gt;And we have a winner! The remainder is zero so x = 1/2 is a root (and (x - 1/2) is a factor). Not only that, the rest of the bottom row tells us the other factor. The &quot;10  6  10  6&quot; translates into {{{10x^3+6x^2+10x+6}}}. We could reset our list of possible roots (using {{{10x^3+6x^2+10x+6}}}) and testing the new list. But I hope that it is apparent that {{{10x^3+6x^2+10x+6}}} will factor by grouping. Since this factoring is easier than the trial method we&#39;ve used so far, that is how I will proceed.&lt;br&gt;
{{{10x^3+6x^2+10x+6}}}
Factor out the GCF of 2:
{{{2(5x^3+3x^2+5x+3)}}}
Then group:
{{{2((5x^3+3x^2)+(5x+3))}}}
Factor out the GCF from each group:
{{{2(x^2(5x+3)+1(5x+3))}}}
(Note: This is one of the rare situations where a GCF of 1 is factored out!) (5x+3) is now a common factor between the groups. So it will factor out:
{{{2(5x+3)(x^2+1)}}}
The factoring is now complete:
{{{10x^4+x^3+7x^2+x-3 = 2(x-1/2)(5x+3)(x^2+1)}}}&lt;br&gt;
Now we can use the Zero Product Property:
{{{2 = 0}}} or {{{x-1/2 = 0}}} or {{{5x+3=0}}} or {{{x^2+1=0}}}
Now we solve these. The first equation is impossible and has no solution. The solution to the second equation is 1/2 (the root we found earlier). The solution to the third equation is -5/3. The fourth equation has complex/imaginary solutions:
{{{x^2+1=0}}}
{{{x^2=-1}}}
x = i or x = -i&lt;br&gt;
This makes the four roots: 1/2, -5/3, i and -i.</PRE>