If `4x^(2)+20x+r=(2x+s)^(2)` for all values of x, what is the values of `r-s`?

Answer

To solve the equation \(4x^2 + 20x + r = (2x + s)^2\) for all values of \(x\) and find the value of \(r - s\), we can follow these steps:

### Step 1: Expand the right side of the equation
We start with the equation:
\[ 4x^2 + 20x + r = (2x + s)^2 \]
Expanding the right side:
\[ (2x + s)^2 = (2x)^2 + 2(2x)(s) + s^2 = 4x^2 + 4sx + s^2 \]
So, we rewrite the equation as:
\[ 4x^2 + 20x + r = 4x^2 + 4sx + s^2 \]

### Step 2: Equate coefficients
Since the equation holds for all values of \(x\), we can equate the coefficients of like terms from both sides.

1. Coefficient of \(x^2\):
\[ 4 = 4 \quad \text{(This is already satisfied)} \]

2. Coefficient of \(x\):
\[ 20 = 4s \]
Solving for \(s\):
\[ s = \frac{20}{4} = 5 \]

3. Constant terms:
\[ r = s^2 \]
Substituting the value of \(s\):
\[ r = 5^2 = 25 \]

### Step 3: Calculate \(r - s\)
Now we can find \(r - s\):
\[ r - s = 25 - 5 = 20 \]

### Final Answer
Thus, the value of \(r - s\) is:
\[ \boxed{20} \]