$$x(x+3)^3-\frac{x}{4}(x+3) = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -3 & x_3 = -\dfrac{ 5 }{ 2 } \\[1 em] x_4 = -\dfrac{ 7 }{ 2 } & \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ 4x^{4}+36x^{3}+107x^{2}+105x = 0 } $, first we need to factor our $ x $.

$$ 4x^{4}+36x^{3}+107x^{2}+105x = x \left( 4x^{3}+36x^{2}+107x+105 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ 4x^{3}+36x^{2}+107x+105 = 0$.

$ \color{blue}{ 4x^{3}+36x^{2}+107x+105 } $ is a polynomial of degree 3. 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 ( 4 ) are 1 2 4 .The factors of the constant term (105) are 1 3 5 7 15 21 35 105 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 2 } , ~ \pm \frac{ 7 }{ 4 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 21 }{ 2 } , ~ \pm \frac{ 21 }{ 4 } , ~ \pm \frac{ 35 }{ 1 } , ~ \pm \frac{ 35 }{ 2 } , ~ \pm \frac{ 35 }{ 4 } , ~ \pm \frac{ 105 }{ 1 } , ~ \pm \frac{ 105 }{ 2 } , ~ \pm \frac{ 105 }{ 4 } ~ $$

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(-3) = 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 + 3} $

$$ \frac{ 4x^{3}+36x^{2}+107x+105 }{ \color{blue}{ x + 3 } } = 4x^{2}+24x+35 $$

Polynomial $ 4x^{2}+24x+35 $ can be used to find the remaining roots.

$ \color{blue}{ 4x^{2}+24x+35 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

Equations Solver