$$(x-1)(x-2)(x-3)(x-4)(x-5) = 0$$

$$ \begin{matrix}x_1 = 1 & x_2 = 2 & x_3 = 3 \\[1 em] x_4 = 4 & x_5 = 5 \\[1 em] \end{matrix} $$

$ \color{blue}{ x^{5}-15x^{4}+85x^{3}-225x^{2}+274x-120 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (-120) are 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 120 }{ 1 } ~ $$

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} $

$$ \frac{ x^{5}-15x^{4}+85x^{3}-225x^{2}+274x-120 }{ \color{blue}{ x - 1 } } = x^{4}-14x^{3}+71x^{2}-154x+120 $$

Polynomial $ x^{4}-14x^{3}+71x^{2}-154x+120 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{4}-14x^{3}+71x^{2}-154x+120 $

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

Polynomial Equations Solver