$$6x^5+23x^4+8x^3-50x^2-40x+8 = 0$$

$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 6 } & x_2 = -2 & x_3 = -\sqrt{ 2 } \\[1 em] x_4 = \sqrt{ 2 } & \\[1 em] \end{matrix} $$

$ \color{blue}{ 6x^{5}+23x^{4}+8x^{3}-50x^{2}-40x+8 } $ is a polynomial of degree 5. 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 ( 6 ) are 1 2 3 6 .The factors of the constant term (8) are 1 2 4 8 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 6 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 3 } , ~ \pm \frac{ 8 }{ 6 } ~ $$

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 }{ 6 }) = 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}{ 6 x - 1 } $

$$ \frac{ 6x^{5}+23x^{4}+8x^{3}-50x^{2}-40x+8 }{ \color{blue}{ 6x - 1 } } = x^{4}+4x^{3}+2x^{2}-8x-8 $$

Polynomial $ x^{4}+4x^{3}+2x^{2}-8x-8 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{4}+4x^{3}+2x^{2}-8x-8 $

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

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