If both `x=2` and `x=(1)/(2)` are factors of `px^(2)+5x+r` , then p=

Options

A. `(3)/(4)r`
B. `2r`
C. `(r )/(2)`
D. r

Answer

To solve the problem, we need to find the value of \( p \) given that both \( x = 2 \) and \( x = \frac{1}{2} \) are factors of the polynomial \( px^2 + 5x + r \).

### Step 1: Use the Factor Theorem
Since \( x = 2 \) is a factor, substituting \( x = 2 \) into the polynomial should yield zero:
\[ p(2)^2 + 5(2) + r = 0 \]
This simplifies to:
\[ 4p + 10 + r = 0 \]
Rearranging gives us our first equation:
\[ 4p + r = -10 \quad \text{(Equation 1)} \]

### Step 2: Substitute the Second Factor
Now, since \( x = \frac{1}{2} \) is also a factor, substituting \( x = \frac{1}{2} \) into the polynomial should also yield zero:
\[ p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0 \]
This simplifies to:
\[ \frac{p}{4} + \frac{5}{2} + r = 0 \]
To eliminate the fraction, multiply the entire equation by 4:
\[ p + 10 + 4r = 0 \]
Rearranging gives us our second equation:
\[ p + 4r = -10 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have two equations:
1. \( 4p + r = -10 \)
2. \( p + 4r = -10 \)

We can solve these equations simultaneously. Let's express \( r \) from Equation 1:
\[ r = -10 - 4p \]
Now substitute this expression for \( r \) into Equation 2:
\[ p + 4(-10 - 4p) = -10 \]
Expanding this gives:
\[ p - 40 - 16p = -10 \]
Combining like terms results in:
\[ -15p - 40 = -10 \]
Adding 40 to both sides:
\[ -15p = 30 \]
Dividing by -15:
\[ p = -2 \]

### Conclusion
Thus, the value of \( p \) is \( -2 \).