$$x^5-13x^4+43x^3-13x^2+42x = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = 6 & x_3 = 7 \\[1 em] x_4 = i & x_5 = -i \\[1 em] \end{matrix} $$

In order to solve $ \color{blue}{ x^{5}-13x^{4}+43x^{3}-13x^{2}+42x = 0 } $, first we need to factor our $ x $.

$$ x^{5}-13x^{4}+43x^{3}-13x^{2}+42x = x \left( x^{4}-13x^{3}+43x^{2}-13x+42 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ x^{4}-13x^{3}+43x^{2}-13x+42 = 0$.

$ \color{blue}{ x^{4}-13x^{3}+43x^{2}-13x+42 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (42) are 1 2 3 6 7 14 21 42 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } ~ $$

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(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}{x - 6} $

$$ \frac{ x^{4}-13x^{3}+43x^{2}-13x+42 }{ \color{blue}{ x - 6 } } = x^{3}-7x^{2}+x-7 $$

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

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

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

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