$$x^4+11x^3+24x^2-23x+35 = 0$$

$$ \begin{matrix}x_1 = -5 & x_2 = -7 & x_3 = \dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 3 }}{ 2 }i \\[1 em] x_4 = \dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 3 }}{ 2 }i & \\[1 em] \end{matrix} $$

$ \color{blue}{ x^{4}+11x^{3}+24x^{2}-23x+35 } $ 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 (35) are 1 5 7 35 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 35 }{ 1 } ~ $$

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(-5) = 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 + 5} $

$$ \frac{ x^{4}+11x^{3}+24x^{2}-23x+35 }{ \color{blue}{ x + 5 } } = x^{3}+6x^{2}-6x+7 $$

Polynomial $ x^{3}+6x^{2}-6x+7 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}+6x^{2}-6x+7 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

Polynomial Equations Solver