$$5x^4-46x^3+84x^2-50x+7 = 0$$

$$ \begin{matrix}x_1 = 1 & x_2 = \dfrac{ 1 }{ 5 } & x_3 = 7 \end{matrix} $$

$ \color{blue}{ 5x^{4}-46x^{3}+84x^{2}-50x+7 } $ 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 ( 5 ) are 1 5 .The factors of the constant term (7) are 1 7 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 5 } ~ $$

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(1) = 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 - 1} $

$$ \frac{ 5x^{4}-46x^{3}+84x^{2}-50x+7 }{ \color{blue}{ x - 1 } } = 5x^{3}-41x^{2}+43x-7 $$

Polynomial $ 5x^{3}-41x^{2}+43x-7 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 5x^{3}-41x^{2}+43x-7 $

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

Polynomial Equations Solver