$$4x^4+5x^3-26x^2-25x+30 = 0$$
Answer
Explanation
$ \color{blue}{ 4x^{4}+5x^{3}-26x^{2}-25x+30 } $ 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 ( 4 ) are 1 2 4 .The factors of the constant term (30) are 1 2 3 5 6 10 15 30 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 10 }{ 4 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 30 }{ 2 } , ~ \pm \frac{ 30 }{ 4 } ~ $$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} $
$$ \frac{ 4x^{4}+5x^{3}-26x^{2}-25x+30 }{ \color{blue}{ x + 2 } } = 4x^{3}-3x^{2}-20x+15 $$Polynomial $ 4x^{3}-3x^{2}-20x+15 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 4x^{3}-3x^{2}-20x+15 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.