Knowing that one root is $x = 1$ means $F(x)$ has a factor $x - 1$. So you can either obtain the complementary factor by long division, or note that:
$\begin{align}F(x) &= x^4(x - 1) + 24x^2(x - 1) - 8x^4 + 23x - 15 \\ &= x^4(x - 1) + 24x^2(x - 1) - 8x(x^3 - 1) + 15(x - 1)\end{align}$
so using $x^3 - 1 = (x - 1)(x^2 + x + 1)$, the complementary factor can be read off as:
$x^4 - 8x^3 + 16x^2 - 8x + 15$
Now if a polynomial with real coefficients has a complex root then it must have the conjugate of that value as another root, and thus having a root $j$ (which I presume denotes $-1^\frac{1}{2}$) means it must have a root $-j$, and thus a factor $(x - j)(x + j)$ which is $x^2 + 1$.
Dividing this out leaves a quadratic factor $x^2 - 8 x + 15$, which by inspection (or solving in the usual way) factors as $(x - 3)(x - 5)$.
edit: DOH! I didn't see yoyo's reply until I had posted mine! May as well leave it now.