$$5x^4-48x^3+171x^2-268x+156 = 0$$
Answer
Explanation
$ \color{blue}{ 5x^{4}-48x^{3}+171x^{2}-268x+156 } $ 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 ( 5 ) are 1 5 .The factors of the constant term (156) are 1 2 3 4 6 12 13 26 39 52 78 156 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 5 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 5 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 5 } , ~ \pm \frac{ 13 }{ 1 } , ~ \pm \frac{ 13 }{ 5 } , ~ \pm \frac{ 26 }{ 1 } , ~ \pm \frac{ 26 }{ 5 } , ~ \pm \frac{ 39 }{ 1 } , ~ \pm \frac{ 39 }{ 5 } , ~ \pm \frac{ 52 }{ 1 } , ~ \pm \frac{ 52 }{ 5 } , ~ \pm \frac{ 78 }{ 1 } , ~ \pm \frac{ 78 }{ 5 } , ~ \pm \frac{ 156 }{ 1 } , ~ \pm \frac{ 156 }{ 5 } ~ $$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} $
$$ \frac{ 5x^{4}-48x^{3}+171x^{2}-268x+156 }{ \color{blue}{ x - 2 } } = 5x^{3}-38x^{2}+95x-78 $$Polynomial $ 5x^{3}-38x^{2}+95x-78 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 5x^{3}-38x^{2}+95x-78 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.