$ \color{blue}{ 4x^{3}-144x^{2}+1280x-2016 } $ is a polynomial of degree 3. 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 (-2016) are 1 2 3 4 6 7 8 9 12 14 16 18 21 24 28 32 36 42 48 56 63 72 84 96 112 126 144 168 224 252 288 336 504 672 1008 2016 . 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{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 2 } , ~ \pm \frac{ 7 }{ 4 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 4 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 4 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 14 }{ 2 } , ~ \pm \frac{ 14 }{ 4 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 16 }{ 4 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 18 }{ 4 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 21 }{ 2 } , ~ \pm \frac{ 21 }{ 4 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 24 }{ 4 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 28 }{ 2 } , ~ \pm \frac{ 28 }{ 4 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 32 }{ 2 } , ~ \pm \frac{ 32 }{ 4 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 36 }{ 4 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 42 }{ 2 } , ~ \pm \frac{ 42 }{ 4 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 48 }{ 4 } , ~ \pm \frac{ 56 }{ 1 } , ~ \pm \frac{ 56 }{ 2 } , ~ \pm \frac{ 56 }{ 4 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 63 }{ 2 } , ~ \pm \frac{ 63 }{ 4 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 72 }{ 2 } , ~ \pm \frac{ 72 }{ 4 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 84 }{ 2 } , ~ \pm \frac{ 84 }{ 4 } , ~ \pm \frac{ 96 }{ 1 } , ~ \pm \frac{ 96 }{ 2 } , ~ \pm \frac{ 96 }{ 4 } , ~ \pm \frac{ 112 }{ 1 } , ~ \pm \frac{ 112 }{ 2 } , ~ \pm \frac{ 112 }{ 4 } , ~ \pm \frac{ 126 }{ 1 } , ~ \pm \frac{ 126 }{ 2 } , ~ \pm \frac{ 126 }{ 4 } , ~ \pm \frac{ 144 }{ 1 } , ~ \pm \frac{ 144 }{ 2 } , ~ \pm \frac{ 144 }{ 4 } , ~ \pm \frac{ 168 }{ 1 } , ~ \pm \frac{ 168 }{ 2 } , ~ \pm \frac{ 168 }{ 4 } , ~ \pm \frac{ 224 }{ 1 } , ~ \pm \frac{ 224 }{ 2 } , ~ \pm \frac{ 224 }{ 4 } , ~ \pm \frac{ 252 }{ 1 } , ~ \pm \frac{ 252 }{ 2 } , ~ \pm \frac{ 252 }{ 4 } , ~ \pm \frac{ 288 }{ 1 } , ~ \pm \frac{ 288 }{ 2 } , ~ \pm \frac{ 288 }{ 4 } , ~ \pm \frac{ 336 }{ 1 } , ~ \pm \frac{ 336 }{ 2 } , ~ \pm \frac{ 336 }{ 4 } , ~ \pm \frac{ 504 }{ 1 } , ~ \pm \frac{ 504 }{ 2 } , ~ \pm \frac{ 504 }{ 4 } , ~ \pm \frac{ 672 }{ 1 } , ~ \pm \frac{ 672 }{ 2 } , ~ \pm \frac{ 672 }{ 4 } , ~ \pm \frac{ 1008 }{ 1 } , ~ \pm \frac{ 1008 }{ 2 } , ~ \pm \frac{ 1008 }{ 4 } , ~ \pm \frac{ 2016 }{ 1 } , ~ \pm \frac{ 2016 }{ 2 } , ~ \pm \frac{ 2016 }{ 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(2) = 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 - 2} $

$$ \frac{ 4x^{3}-144x^{2}+1280x-2016 }{ \color{blue}{ x - 2 } } = 4x^{2}-136x+1008 $$

Polynomial $ 4x^{2}-136x+1008 $ can be used to find the remaining roots.

$ \color{blue}{ 4x^{2}-136x+1008 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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