Let `p(x)` be a polynomial of degree 4 such that `P(1)=P(3)=P(5)=P'(7)=0.` If the real number `ane1,3,5` is such that `P(a)=0` can be expressed as `a=(p)/(q)`, where p and q are relatively prime, then `(p-8q) is……….

Answer

To solve the problem step by step, we start by analyzing the given conditions for the polynomial \( P(x) \).

### Step 1: Formulate the Polynomial
Since \( P(x) \) is a polynomial of degree 4 and has roots at \( x = 1, 3, 5 \), we can express it as:
\[ P(x) = k(x - 1)(x - 3)(x - 5)(x - \alpha) \]
where \( k \) is a constant and \( \alpha \) is another root we need to find.

### Step 2: Use the Derivative Condition
We know that \( P'(7) = 0 \). To find \( P'(x) \), we can use the product rule:
\[ P'(x) = k \left[ (x - 3)(x - 5)(x - \alpha) + (x - 1)(x - 5)(x - \alpha) + (x - 1)(x - 3)(x - \alpha) + (x - 1)(x - 3)(x - 5) \right] \]

### Step 3: Substitute \( x = 7 \)
Now, we substitute \( x = 7 \) into \( P'(x) \):
\[ P'(7) = k \left[ (7 - 3)(7 - 5)(7 - \alpha) + (7 - 1)(7 - 5)(7 - \alpha) + (7 - 1)(7 - 3)(7 - \alpha) + (7 - 1)(7 - 3)(7 - 5) \right] \]
This simplifies to:
\[ P'(7) = k \left[ 4(2)(7 - \alpha) + 6(2)(7 - \alpha) + 6(4)(2) \right] \]
Setting this equal to zero gives us:
\[ 4(2)(7 - \alpha) + 6(2)(7 - \alpha) + 48 = 0 \]

### Step 4: Solve for \( \alpha \)
Combining like terms:
\[ (8 + 12)(7 - \alpha) + 48 = 0 \]
\[ 20(7 - \alpha) + 48 = 0 \]
\[ 20(7 - \alpha) = -48 \]
\[ 7 - \alpha = -\frac{48}{20} = -\frac{12}{5} \]
Thus,
\[ \alpha = 7 + \frac{12}{5} = \frac{35}{5} + \frac{12}{5} = \frac{47}{5} \]

### Step 5: Express \( \alpha \) in the Form \( \frac{p}{q} \)
Here, \( \alpha = \frac{47}{5} \), where \( p = 47 \) and \( q = 5 \). Since 47 and 5 are relatively prime, we can proceed to the next step.

### Step 6: Calculate \( p - 8q \)
Now we compute:
\[ p - 8q = 47 - 8 \times 5 = 47 - 40 = 7 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{7} \]