Given the trinomial x^2+bx-c where both the first sign is positive and the second sign is negative, the signs of the factors will be:

The signs of the factors will be different. (One will be positive and one will be negative.)

Here, the "first sign" is considered to be the sign of x², and the "second sign" is considered to be the sign of c. The above statement will be true regardless of the sign of b. The "signs of the factors" is considered to refer to the signs of the constant terms in the binomial factors.

For (x +p)(x -q) where p and q are both positive so the signs are as shown, the product is x² +(p-q)x -pq. That is x² is positive and -pq is negative, regardless of the relative magnitudes of p and q (thus the sign of p-q).

If something else is meant by the terminology used, it isn't clear what the intended answer is supposed to be.

Both x² +3x -10 and x² -3x -10 have factorizations that have different signs in the two factors. The first is (x+5)(x-2); the second is (x+2)(x-5). The signs of the factors are all positive: (x+5), not -(x+5), for example.

On the other hand x² -3x +2 = (x-2)(x-1) has same signs in and of the factors. That is, by themselves, the sign of x² (first sign) and the sign of 3x (second sign) don't guarantee the signs in the factors are anything in particular, except that at least one sign in the factors must be negative if the first (x²) and second (3x) signs differ.

r and t.

both are 64 degrees.