In order to solve $ \color{blue}{ 7x^{4}-1512x = 0 } $, first we need to factor our $ x $.

$$ 7x^{4}-1512x = x \left( 7x^{3}-1512 \right) $$

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

The remaining roots can be found by solving equation $ 7x^{3}-1512 = 0$.

$ \color{blue}{ 7x^{3}-1512 } $ 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 ( 7 ) are 1 7 .The factors of the constant term (-1512) are 1 2 3 4 6 7 8 9 12 14 18 21 24 27 28 36 42 54 56 63 72 84 108 126 168 189 216 252 378 504 756 1512 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 7 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 7 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 7 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 7 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 7 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 7 }{ 7 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 7 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 7 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 7 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 14 }{ 7 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 7 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 21 }{ 7 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 7 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 7 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 28 }{ 7 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 7 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 42 }{ 7 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 7 } , ~ \pm \frac{ 56 }{ 1 } , ~ \pm \frac{ 56 }{ 7 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 63 }{ 7 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 72 }{ 7 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 84 }{ 7 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 108 }{ 7 } , ~ \pm \frac{ 126 }{ 1 } , ~ \pm \frac{ 126 }{ 7 } , ~ \pm \frac{ 168 }{ 1 } , ~ \pm \frac{ 168 }{ 7 } , ~ \pm \frac{ 189 }{ 1 } , ~ \pm \frac{ 189 }{ 7 } , ~ \pm \frac{ 216 }{ 1 } , ~ \pm \frac{ 216 }{ 7 } , ~ \pm \frac{ 252 }{ 1 } , ~ \pm \frac{ 252 }{ 7 } , ~ \pm \frac{ 378 }{ 1 } , ~ \pm \frac{ 378 }{ 7 } , ~ \pm \frac{ 504 }{ 1 } , ~ \pm \frac{ 504 }{ 7 } , ~ \pm \frac{ 756 }{ 1 } , ~ \pm \frac{ 756 }{ 7 } , ~ \pm \frac{ 1512 }{ 1 } , ~ \pm \frac{ 1512 }{ 7 } ~ $$

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} $

$$ \frac{ 7x^{3}-1512 }{ \color{blue}{ x - 6 } } = 7x^{2}+42x+252 $$

Polynomial $ 7x^{2}+42x+252 $ can be used to find the remaining roots.

$ \color{blue}{ 7x^{2}+42x+252 } $ 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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