$$(3-2x)(6x+4)\cdot(5-8x) = 0$$

$$ \begin{matrix}x_1 = -\dfrac{ 2 }{ 3 } & x_2 = \dfrac{ 3 }{ 2 } & x_3 = \dfrac{ 5 }{ 8 } \end{matrix} $$

$ \color{blue}{ 96x^{3}-140x^{2}-46x+60 } $ 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 ( 96 ) are 1 2 3 4 6 8 12 16 24 32 48 96 .The factors of the constant term (60) are 1 2 3 4 5 6 10 12 15 20 30 60 . 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 }{ 4 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 12 } , ~ \pm \frac{ 1 }{ 16 } , ~ \pm \frac{ 1 }{ 24 } , ~ \pm \frac{ 1 }{ 32 } , ~ \pm \frac{ 1 }{ 48 } , ~ \pm \frac{ 1 }{ 96 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 2 }{ 8 } , ~ \pm \frac{ 2 }{ 12 } , ~ \pm \frac{ 2 }{ 16 } , ~ \pm \frac{ 2 }{ 24 } , ~ \pm \frac{ 2 }{ 32 } , ~ \pm \frac{ 2 }{ 48 } , ~ \pm \frac{ 2 }{ 96 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 3 }{ 12 } , ~ \pm \frac{ 3 }{ 16 } , ~ \pm \frac{ 3 }{ 24 } , ~ \pm \frac{ 3 }{ 32 } , ~ \pm \frac{ 3 }{ 48 } , ~ \pm \frac{ 3 }{ 96 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 4 }{ 6 } , ~ \pm \frac{ 4 }{ 8 } , ~ \pm \frac{ 4 }{ 12 } , ~ \pm \frac{ 4 }{ 16 } , ~ \pm \frac{ 4 }{ 24 } , ~ \pm \frac{ 4 }{ 32 } , ~ \pm \frac{ 4 }{ 48 } , ~ \pm \frac{ 4 }{ 96 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 3 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 5 }{ 6 } , ~ \pm \frac{ 5 }{ 8 } , ~ \pm \frac{ 5 }{ 12 } , ~ \pm \frac{ 5 }{ 16 } , ~ \pm \frac{ 5 }{ 24 } , ~ \pm \frac{ 5 }{ 32 } , ~ \pm \frac{ 5 }{ 48 } , ~ \pm \frac{ 5 }{ 96 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 6 }{ 6 } , ~ \pm \frac{ 6 }{ 8 } , ~ \pm \frac{ 6 }{ 12 } , ~ \pm \frac{ 6 }{ 16 } , ~ \pm \frac{ 6 }{ 24 } , ~ \pm \frac{ 6 }{ 32 } , ~ \pm \frac{ 6 }{ 48 } , ~ \pm \frac{ 6 }{ 96 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 10 }{ 3 } , ~ \pm \frac{ 10 }{ 4 } , ~ \pm \frac{ 10 }{ 6 } , ~ \pm \frac{ 10 }{ 8 } , ~ \pm \frac{ 10 }{ 12 } , ~ \pm \frac{ 10 }{ 16 } , ~ \pm \frac{ 10 }{ 24 } , ~ \pm \frac{ 10 }{ 32 } , ~ \pm \frac{ 10 }{ 48 } , ~ \pm \frac{ 10 }{ 96 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 3 } , ~ \pm \frac{ 12 }{ 4 } , ~ \pm \frac{ 12 }{ 6 } , ~ \pm \frac{ 12 }{ 8 } , ~ \pm \frac{ 12 }{ 12 } , ~ \pm \frac{ 12 }{ 16 } , ~ \pm \frac{ 12 }{ 24 } , ~ \pm \frac{ 12 }{ 32 } , ~ \pm \frac{ 12 }{ 48 } , ~ \pm \frac{ 12 }{ 96 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 3 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 15 }{ 6 } , ~ \pm \frac{ 15 }{ 8 } , ~ \pm \frac{ 15 }{ 12 } , ~ \pm \frac{ 15 }{ 16 } , ~ \pm \frac{ 15 }{ 24 } , ~ \pm \frac{ 15 }{ 32 } , ~ \pm \frac{ 15 }{ 48 } , ~ \pm \frac{ 15 }{ 96 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 20 }{ 2 } , ~ \pm \frac{ 20 }{ 3 } , ~ \pm \frac{ 20 }{ 4 } , ~ \pm \frac{ 20 }{ 6 } , ~ \pm \frac{ 20 }{ 8 } , ~ \pm \frac{ 20 }{ 12 } , ~ \pm \frac{ 20 }{ 16 } , ~ \pm \frac{ 20 }{ 24 } , ~ \pm \frac{ 20 }{ 32 } , ~ \pm \frac{ 20 }{ 48 } , ~ \pm \frac{ 20 }{ 96 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 30 }{ 2 } , ~ \pm \frac{ 30 }{ 3 } , ~ \pm \frac{ 30 }{ 4 } , ~ \pm \frac{ 30 }{ 6 } , ~ \pm \frac{ 30 }{ 8 } , ~ \pm \frac{ 30 }{ 12 } , ~ \pm \frac{ 30 }{ 16 } , ~ \pm \frac{ 30 }{ 24 } , ~ \pm \frac{ 30 }{ 32 } , ~ \pm \frac{ 30 }{ 48 } , ~ \pm \frac{ 30 }{ 96 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 60 }{ 2 } , ~ \pm \frac{ 60 }{ 3 } , ~ \pm \frac{ 60 }{ 4 } , ~ \pm \frac{ 60 }{ 6 } , ~ \pm \frac{ 60 }{ 8 } , ~ \pm \frac{ 60 }{ 12 } , ~ \pm \frac{ 60 }{ 16 } , ~ \pm \frac{ 60 }{ 24 } , ~ \pm \frac{ 60 }{ 32 } , ~ \pm \frac{ 60 }{ 48 } , ~ \pm \frac{ 60 }{ 96 } ~ $$

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(-\frac{ 2 }{ 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}{ 3 x + 2 } $

$$ \frac{ 96x^{3}-140x^{2}-46x+60 }{ \color{blue}{ 3x + 2 } } = 32x^{2}-68x+30 $$

Polynomial $ 32x^{2}-68x+30 $ can be used to find the remaining roots.

$ \color{blue}{ 32x^{2}-68x+30 } $ 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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