$$(x+3)(x+5)(x+7)(x+9) = 0$$
$$ \begin{matrix}x_1 = -3 & x_2 = -5 & x_3 = -7 \\[1 em] x_4 = -9 & \\[1 em] \end{matrix} $$
$ \color{blue}{ x^{4}+24x^{3}+206x^{2}+744x+945 } $ 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 (945) are 1 3 5 7 9 15 21 27 35 45 63 105 135 189 315 945 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 35 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 105 }{ 1 } , ~ \pm \frac{ 135 }{ 1 } , ~ \pm \frac{ 189 }{ 1 } , ~ \pm \frac{ 315 }{ 1 } , ~ \pm \frac{ 945 }{ 1 } ~ $$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} $
$$ \frac{ x^{4}+24x^{3}+206x^{2}+744x+945 }{ \color{blue}{ x + 3 } } = x^{3}+21x^{2}+143x+315 $$Polynomial $ x^{3}+21x^{2}+143x+315 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+21x^{2}+143x+315 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.