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

Options

A. `-3`
B. `-2`
C. `2`
D. `8`

Explanation

To solve the equation \( x^2 - 2x - 15 = (x + r)(x + s) \) for all values of \( x \) and find a possible value of \( r - s \), we can follow these steps:

### Step 1: Factor the left-hand side of the equation
We start with the quadratic expression on the left side:
\[ x^2 - 2x - 15 \]

We need to factor this expression. We are looking for two numbers that multiply to \(-15\) (the constant term) and add up to \(-2\) (the coefficient of \(x\)).

The factors of \(-15\) that add up to \(-2\) are:
- \(3\) and \(-5\)

So we can rewrite the quadratic as:
\[ x^2 - 5x + 3x - 15 \]

### Step 2: Group the terms and factor by grouping
Now we group the terms:
\[ (x^2 - 5x) + (3x - 15) \]

Factoring out the common terms in each group gives us:
\[ x(x - 5) + 3(x - 5) \]

Now we can factor out \((x - 5)\):
\[ (x - 5)(x + 3) \]

### Step 3: Set the factored form equal to the right-hand side
Now we have:
\[ x^2 - 2x - 15 = (x - 5)(x + 3) \]

We can compare this with the right-hand side of the original equation:
\[ (x + r)(x + s) \]

From our factorization, we can identify:
- \(r = 3\)
- \(s = -5\)

### Step 4: Calculate \(r - s\)
Now we need to find \(r - s\):
\[ r - s = 3 - (-5) = 3 + 5 = 8 \]

### Conclusion
Thus, one possible value of \(r - s\) is:
\[ \boxed{8} \]

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