$$x^4-7x^3-5x^2+75x = 0$$
Answer
Explanation
In order to solve $ \color{blue}{ x^{4}-7x^{3}-5x^{2}+75x = 0 } $, first we need to factor our $ x $.
$$ x^{4}-7x^{3}-5x^{2}+75x = x \left( x^{3}-7x^{2}-5x+75 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{3}-7x^{2}-5x+75 = 0$.
$ \color{blue}{ x^{3}-7x^{2}-5x+75 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (75) are 1 3 5 15 25 75 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } ~ $$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} $
$$ \frac{ x^{3}-7x^{2}-5x+75 }{ \color{blue}{ x + 3 } } = x^{2}-10x+25 $$Polynomial $ x^{2}-10x+25 $ can be used to find the remaining roots.
$ \color{blue}{ x^{2}-10x+25 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.