$ \color{blue}{ -10x^{3}-11x^{2}+379x+38 } $ 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 ( -10 ) are 1 2 5 10 .The factors of the constant term (38) are 1 2 19 38 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 2 }{ 10 } , ~ \pm \frac{ 19 }{ 1 } , ~ \pm \frac{ 19 }{ 2 } , ~ \pm \frac{ 19 }{ 5 } , ~ \pm \frac{ 19 }{ 10 } , ~ \pm \frac{ 38 }{ 1 } , ~ \pm \frac{ 38 }{ 2 } , ~ \pm \frac{ 38 }{ 5 } , ~ \pm \frac{ 38 }{ 10 } ~ $$

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{ 1 }{ 10 }) = 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}{ 10 x + 1 } $

$$ \frac{ -10x^{3}-11x^{2}+379x+38 }{ \color{blue}{ 10x + 1 } } = -x^{2}-x+38 $$

Polynomial $ -x^{2}-x+38 $ can be used to find the remaining roots.

$ \color{blue}{ -x^{2}-x+38 } $ 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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