$$x^4-\frac{41}{2}x^3+148x^2-438x+432 = 0$$

$$ \begin{matrix}x_1 = 2 & x_2 = 6 & x_3 = 8 \\[1 em] x_4 = \dfrac{ 9 }{ 2 } & \\[1 em] \end{matrix} $$

$ \color{blue}{ 2x^{4}-41x^{3}+296x^{2}-876x+864 } $ is a polynomial of degree 4. 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 (864) are 1 2 3 4 6 8 9 12 16 18 24 27 32 36 48 54 72 96 108 144 216 288 432 864 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 32 }{ 2 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 72 }{ 2 } , ~ \pm \frac{ 96 }{ 1 } , ~ \pm \frac{ 96 }{ 2 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 108 }{ 2 } , ~ \pm \frac{ 144 }{ 1 } , ~ \pm \frac{ 144 }{ 2 } , ~ \pm \frac{ 216 }{ 1 } , ~ \pm \frac{ 216 }{ 2 } , ~ \pm \frac{ 288 }{ 1 } , ~ \pm \frac{ 288 }{ 2 } , ~ \pm \frac{ 432 }{ 1 } , ~ \pm \frac{ 432 }{ 2 } , ~ \pm \frac{ 864 }{ 1 } , ~ \pm \frac{ 864 }{ 2 } ~ $$

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} $

$$ \frac{ 2x^{4}-41x^{3}+296x^{2}-876x+864 }{ \color{blue}{ x - 2 } } = 2x^{3}-37x^{2}+222x-432 $$

Polynomial $ 2x^{3}-37x^{2}+222x-432 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 2x^{3}-37x^{2}+222x-432 $

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

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