$$(2x+4)(x-7)(x+3) = 0$$

$$ \begin{matrix}x_1 = -2 & x_2 = -3 & x_3 = 7 \end{matrix} $$

$ \color{blue}{ 2x^{3}-4x^{2}-58x-84 } $ 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 ( 2 ) are 1 2 .The factors of the constant term (-84) are 1 2 3 4 6 7 12 14 21 28 42 84 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 2 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 14 }{ 2 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 21 }{ 2 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 28 }{ 2 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 42 }{ 2 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 84 }{ 2 } ~ $$

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} $

$$ \frac{ 2x^{3}-4x^{2}-58x-84 }{ \color{blue}{ x + 2 } } = 2x^{2}-8x-42 $$

Polynomial $ 2x^{2}-8x-42 $ can be used to find the remaining roots.

$ \color{blue}{ 2x^{2}-8x-42 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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