$$8 \cdot \frac{x}{2x+4}+9x-\frac{15}{3x+6} = 6x-\frac{7}{x+2}$$

$$ \begin{matrix}x_1 = -2 & x_2 = -\dfrac{ 5 }{ 3 }-\dfrac{\sqrt{ 19 }}{ 3 } & x_3 = -\dfrac{ 5 }{ 3 }+\dfrac{\sqrt{ 19 }}{ 3 } \end{matrix} $$

$ \color{blue}{ 18x^{4}+132x^{3}+324x^{2}+288x+48 } $ 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 ( 18 ) are 1 2 3 6 9 18 .The factors of the constant term (48) are 1 2 3 4 6 8 12 16 24 48 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 1 }{ 18 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 2 }{ 9 } , ~ \pm \frac{ 2 }{ 18 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 3 }{ 9 } , ~ \pm \frac{ 3 }{ 18 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 6 } , ~ \pm \frac{ 4 }{ 9 } , ~ \pm \frac{ 4 }{ 18 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 6 }{ 6 } , ~ \pm \frac{ 6 }{ 9 } , ~ \pm \frac{ 6 }{ 18 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 3 } , ~ \pm \frac{ 8 }{ 6 } , ~ \pm \frac{ 8 }{ 9 } , ~ \pm \frac{ 8 }{ 18 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 3 } , ~ \pm \frac{ 12 }{ 6 } , ~ \pm \frac{ 12 }{ 9 } , ~ \pm \frac{ 12 }{ 18 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 16 }{ 3 } , ~ \pm \frac{ 16 }{ 6 } , ~ \pm \frac{ 16 }{ 9 } , ~ \pm \frac{ 16 }{ 18 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 24 }{ 3 } , ~ \pm \frac{ 24 }{ 6 } , ~ \pm \frac{ 24 }{ 9 } , ~ \pm \frac{ 24 }{ 18 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 48 }{ 3 } , ~ \pm \frac{ 48 }{ 6 } , ~ \pm \frac{ 48 }{ 9 } , ~ \pm \frac{ 48 }{ 18 } ~ $$

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} $

$$ \frac{ 18x^{4}+132x^{3}+324x^{2}+288x+48 }{ \color{blue}{ x + 2 } } = 18x^{3}+96x^{2}+132x+24 $$

Polynomial $ 18x^{3}+96x^{2}+132x+24 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 18x^{3}+96x^{2}+132x+24 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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