factor the following polynomials completely -3x^2-10s+8 now would i find the middle factor...?

product of 8 and -3 = -24
factors for -24 are 1)-12 ,2
2) 12, -2
3) 4 , -6
4) -4 , 6
5) 8, -3
6) 3, -8

Of the above only the factors 1), that is -12, 2 when added give -10.

Therefore we would take those factors. And split the middle term to get the following:

-3x^2 -12x + 2x + 8

Now factorize to get: -3x( x + 4 ) + 2 ( x + 4 )
= (-3x + 2)( x + 4)

I assume you meant -10x, not -10s. So, you want to factor
-3x^2 -10x + 8. First, factor the minus sign and you get
-(3x^2 + 10x - 8). Now factor inside the parenthesis and you get[3(x+4)(x-2/3)]. You can get rid of the 2/3 by multiplying through by the 3 term to get (3x-2). So, you get -(3x-2)(x+4). A useful thing for factoring quadratic equations is this. Treat it as an equation and set it = 0.
you will get to values, call them r1 and r2. In ANY polynomial p(x), if there exists a value r such that
p(r)=0, then x-r is always a factor. When I used the quadratic equation, I got x = -4, and x = 2/3, so that means x + 4 and x - 2/3 are factors. Let me know if this is not clear.

Do you mean -3x^2-10x+8 ???
The answer should be (-3x+2)(x+4)
x=2/3 or x=-4