How can you determine all of the zeroes (real and

Step 1
You can expect any factors to divide -24, the constant term over the lead coefficient. So check 1, 2, 3, 4, 6,12, 24 and their negatives. Turns out that 3 and -4 work, so you are down to a quadratic
$P(x)=(x-3)\times (x+4)\times (x^2+x+2)$
so the roots are
$x=\{3,-4,\frac{-1-\sqrt{7}\times i}{2},\frac{-1+\sqrt{7}\times i}{2}\}$

Step 1
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
$x^4+2x^3-9x^2-10x-24=(x^2+x+k{)}^2-(mx+n{)}^2$
that is:
$x^4+2x^3+(1+2k-m^2)x^2+2(k-mn)x+(k^2-n^2)=x^4+2x^3-9x^2-10x-24$
The basic trick is to try finding 3 numbers k,m,n which would enable us to write:
$$\begin{gathered} 1+2k-m^2=-9 \\ 2k-2mn=-10 \\ {k}^2-n^2=-24 \end{gathered}$$
Rearrange these equations and get
$$\begin{gathered} m^2=2(k+5) \\ mn=k+5 \\ {n}^2=k^2+24 \end{gathered}$$
From the 2 equations on the left we get
$m^2=2mn$
and therefore one possible solution would be:
$$\begin{gathered} m=0 \\ k=-5 \\ {n}^2=49=7^2 \end{gathered}$$
Hence we get the following factorization of the given polynomial:
$$\begin{gathered} x^4+2x^3-9x^2-10x-24=(x^2+x-5{)}^2-7^2= \\ =(x^2+x+2)(x^2+x-12) \end{gathered}$$
So, all the problem has now been reduced into finding the 4 roots of the 2 quadratic polynomials, which are
$$\begin{gathered} x_1=-\frac{1}{2}+\frac{\sqrt{7}}{2}i \\ {x}_2=-\frac{1}{2}-\frac{\sqrt{7}}{2}i \end{gathered}$$
and
$$\begin{gathered} x_3=-\frac{1}{2}+\frac{7}{2}=3 \\ {x}_4=-\frac{1}{2}-\frac{7}{2}=-4 \end{gathered}$$