$$4(x-1)^2(x^2+5) = 0$$

$$ \begin{matrix}x_1 = 1 & x_2 = \sqrt{ 5 } i & x_3 = -\sqrt{ 5 } i \end{matrix} $$

$ \color{blue}{ 4x^{4}-8x^{3}+24x^{2}-40x+20 } $ 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 ( 4 ) are 1 2 4 .The factors of the constant term (20) are 1 2 4 5 10 20 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 10 }{ 4 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 20 }{ 2 } , ~ \pm \frac{ 20 }{ 4 } ~ $$

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} $

$$ \frac{ 4x^{4}-8x^{3}+24x^{2}-40x+20 }{ \color{blue}{ x - 1 } } = 4x^{3}-4x^{2}+20x-20 $$

Polynomial $ 4x^{3}-4x^{2}+20x-20 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 4x^{3}-4x^{2}+20x-20 $

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

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