$$x^4-15x^3+62x^2-90x+336 = 0$$
Answer
Explanation
$ \color{blue}{ x^{4}-15x^{3}+62x^{2}-90x+336 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (336) are 1 2 3 4 6 7 8 12 14 16 21 24 28 42 48 56 84 112 168 336 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 56 }{ 1 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 112 }{ 1 } , ~ \pm \frac{ 168 }{ 1 } , ~ \pm \frac{ 336 }{ 1 } ~ $$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(7) = 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 - 7} $
$$ \frac{ x^{4}-15x^{3}+62x^{2}-90x+336 }{ \color{blue}{ x - 7 } } = x^{3}-8x^{2}+6x-48 $$Polynomial $ x^{3}-8x^{2}+6x-48 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}-8x^{2}+6x-48 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.