Question 935110
1) using synthetic division, we divide x^4+px^2+qx-40 by (x+2), note that the equation becomes x^4+0x^3+px^2+qx-40, then we have
-2 |  1  0  p q -40  |
we get the following cubic x^3-x^2+(4+p)x-8-2p+q where -8-2p+q is the constant
we also know that 16+4p-2q-40 = 0, therefore 4p-2q = 24, 2p-q=12
2) now use the cubic result and divide it by synthetic division
5 | 1  -2  4+p  -8-2p+q
we get the following quadratic x^2+3x+19+p where 19+p is the constant
we also know that 95+5p-8-2p+q=0, 3p+q=-87
our two equations are
2p-q=12
3p+q=-87
solve first equation for p
p = (q/2)+6
substitute for p in second equation and solve for q, then p
3((q/2)+6)+q = -87
multiply both sides of = by 2
3q+36+2q=-174
5q = -210
q = -42
p = -15
our original equation is
x^4+px^2+qx-40
substitute for p and q
x^4-15x^2-42x-40
zeros for this equation using its graph
{{{ graph( 300, 200, -5, 6, -300, 600, x^4-15x^2-42x-40 ) }}} 
we see the graph crosses the x-axis at -2 and 5, these are the only zeros