Polynomial equation solver

Solve equation$$-x^5+12x^4-36x^3-54x^2+405x-486 = 0$$

solution

$$ \begin{matrix}x_1 = 3 & x_2 = -3 & x_3 = 6 \end{matrix} $$

explanation

$ \color{blue}{ -x^{5}+12x^{4}-36x^{3}-54x^{2}+405x-486 } $ 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 (-486) are 1 2 3 6 9 18 27 54 81 162 243 486 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 81 }{ 1 } , ~ \pm \frac{ 162 }{ 1 } , ~ \pm \frac{ 243 }{ 1 } , ~ \pm \frac{ 486 }{ 1 } ~ $$

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} $

$$ \frac{ -x^{5}+12x^{4}-36x^{3}-54x^{2}+405x-486 }{ \color{blue}{ x - 3 } } = -x^{4}+9x^{3}-9x^{2}-81x+162 $$

Polynomial $ -x^{4}+9x^{3}-9x^{2}-81x+162 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ -x^{4}+9x^{3}-9x^{2}-81x+162 $

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