if x^2+2x-15=(x+r)(x+s) for all values of x one possible value of for r-s is?

x^2+2x-15=(x+r)(x+s)

(x+5)(x-3) = (x+r)(x+s)

so r = 5 and s = -3

r - s
= 5 - (-3)
= 8

To determine the value of r - s, we can compare the given equation with the factored form of the quadratic equation.

The factored form of a quadratic equation of the form ax^2 + bx + c is (x - p)(x - q), where p and q are the solutions or roots of the equation.

In the given equation x^2 + 2x - 15 = (x + r)(x + s), we can see that the coefficients of x are the same, so the x terms in the factored form should add up to 2x.

Expanding the factored form (x + r)(x + s) gives us x^2 + (r + s)x + rs.

Comparing this with the given equation, we can equate the coefficients:

r + s = 2 (from the coefficient of x)
rs = -15 (from the constant term)

Now, to find a possible value for r - s, we can solve these two equations simultaneously.

First, let's consider the possible values for r and s that satisfy the second equation rs = -15:

- r = -1, s = 15
- r = -3, s = 5
- r = 1, s = -15
- r = 3, s = -5

Now, substituting these values into the first equation r + s = 2, we can find the values of r - s:

- (r - s) = 2 - (-1) = 3
- (r - s) = 2 - (-3) = 5
- (r - s) = 2 - 1 = 1
- (r - s) = 2 - 3 = -1

Therefore, the possible value for r - s is 3, 5, 1, or -1, depending on the values chosen for r and s.