Question 389368
(Note: I am assuming that there is a minus in front of the 2x because the problem is much more difficult if there is a plus.)
{{{x^4+ 8x^3 + 15x^2 -  2x - 10 = 0}}}
To find solutions to this equation we will need to factor it. The only way to factor this, as far as I can see, is to use trial and error of the possible rational roots.<br>
The possible rational roots of a polynomial are all the possible fractions, positive and negative, that can be formed using a factor of the constant term (at the end) over a factor of the leading coefficient (in front of the term with the highest exponent). Your constant term is -10 and you leading coefficient is 1, So the list of possible rational roots are:
10/1, -10/1, 5/1, -5/1, 2/1, -2/1, 1/1 and -1/1
which simplify to:
10, -10, 5, -5, 2, -2, 1 and -1<br>
Synthetic division is often the method used to check these roots. This is a trial and error method but I will not waste time showing the ones that don't work. First we will try -5:
<pre>
-5 |   1   8   15  -2   -10
----      -5  -15   0    10
      ---------------------
       1   3    0  -2     0
</pre>
Above we have divided {{{x^4+ 8x^3 + 15x^2 -  2x - 10}}} by (x - (-5)) (or (x+5)). The zero in the lower right corner is the remainder of this division. This means (x+5) divides evenly. This means that (x+5) is a factor of {{{x^4+ 8x^3 + 15x^2 -  2x - 10}}}. The rest of the bottom row tells us the other factor. The 1 3 0 -2 translates into {{{x^3 + 3x^2 - 2}}}.<br>
Our  equation is now:
{{{(x+5)(x^3 + 3x^2 - 2) = 0}}}
We now look for factors of {{{x^3 + 3x^2 - 2}}}. Its possible rational roots are: 2, -2, 1 and -1. -1 works:
<pre>
-1 |  1   3   0   -2
----     -1  -2    2
     ---------------
      1   2  -2    0
</pre>
So (x - (-1)) or (x+1) is a factor of {{{x^3 + 3x^2 - 2}}} and the other factor (1 2 -2) is {{{x^2 + 2x - 2}}}. Our equation is now:
{{{(x+5)(x+1)(x^2+2x-2) = 0}}}<br>
The third factor is a quadratic which will not factor easily. But since it is a quadratic we can use the Quadratic Formula to find the values of x that make it zero. So we can proceed to the next step.<br>
The Zero Product Property tells us that this (or any) product can be zero <i>only</i> if one (or more) of the factors is zero. So:
x+5 = 0 or x+1 = 0 or {{{x^2+3x-2 = 0}}}
The first two are simple to solve. We get x = -5 and x = -1 for solutions. The third equation requires use of the Quadratic Formula:
{{{x = (-(2) +- sqrt((2)^2 - 4(1)(-2)))/2(1)}}}
which simplifies as follows:
{{{x = (-(2) +- sqrt(4 - 4(1)(-2)))/2(1)}}}
{{{x = (-(2) +- sqrt(4 + 8))/2(1)}}}
{{{x = (-(2) +- sqrt(12))/2(1)}}}
{{{x = (-2 +- sqrt(12))/2}}}
{{{x = (-2 +- sqrt(4*3))/2}}}
{{{x = (-2 +- sqrt(4)*sqrt(3))/2}}}
{{{x = (-2 +- 2*sqrt(3))/2}}}
{{{x = (2(-1 +- sqrt(3)))/2}}}
{{{x = (cross(2)(-1 +- sqrt(3)))/cross(2)}}}
{{{x = -1 +- sqrt(3)}}}
In long form this is:
{{{x = -1 + sqrt(3)}}} or {{{x = -1 - sqrt(3)}}}<br>
So the solutions to your equation are:
x = -5 or x = -1 or {{{x = -1 + sqrt(3)}}} or {{{x = -1 - sqrt(3)}}}