$$\frac{2}{(x+6)(x-4)}-\frac{1}{(x+6)(x+2)} = \frac{3}{(x-4)(x-2)}$$
Answer
Explanation
$ \color{blue}{ -2x^{4}-26x^{3}+84x^{2}+104x-304 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.
The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.
The factors of the leading coefficient ( -2 ) are 1 2 .The factors of the constant term (-304) are 1 2 4 8 16 19 38 76 152 304 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 19 }{ 1 } , ~ \pm \frac{ 19 }{ 2 } , ~ \pm \frac{ 38 }{ 1 } , ~ \pm \frac{ 38 }{ 2 } , ~ \pm \frac{ 76 }{ 1 } , ~ \pm \frac{ 76 }{ 2 } , ~ \pm \frac{ 152 }{ 1 } , ~ \pm \frac{ 152 }{ 2 } , ~ \pm \frac{ 304 }{ 1 } , ~ \pm \frac{ 304 }{ 2 } ~ $$Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
If we plug these values into the polynomial $ P(x) $, we obtain $ P(2) = 0 $.
To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:
Divide $ P(x) $ with $ \color{blue}{x - 2} $
$$ \frac{ -2x^{4}-26x^{3}+84x^{2}+104x-304 }{ \color{blue}{ x - 2 } } = -2x^{3}-30x^{2}+24x+152 $$Polynomial $ -2x^{3}-30x^{2}+24x+152 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ -2x^{3}-30x^{2}+24x+152 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.