$$x^4-2x^3+x^2+18x-90 = 0$$

$$ \begin{matrix}x_1 = 3 & x_2 = -3 & x_3 = 1+3i \\[1 em] x_4 = 1-3i & \\[1 em] \end{matrix} $$

$ \color{blue}{ x^{4}-2x^{3}+x^{2}+18x-90 } $ 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 (-90) are 1 2 3 5 6 9 10 15 18 30 45 90 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } ~ $$

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(3) = 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 - 3} $

$$ \frac{ x^{4}-2x^{3}+x^{2}+18x-90 }{ \color{blue}{ x - 3 } } = x^{3}+x^{2}+4x+30 $$

Polynomial $ x^{3}+x^{2}+4x+30 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}+x^{2}+4x+30 $

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

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