For what value of p, the equation `(p-1)(2p+1)xx^2+(p^2-1)xx+(p-1)(p-3)=0` is an identity (or hs more than two solutions)?

Options

A. `-1`
B. `-1//2`
C. 1
D. `1//3`

Answer

To determine the value of \( p \) for which the equation

\[ (p-1)(2p+1)x^2 + (p^2-1)x + (p-1)(p-3) = 0 \]

is an identity (true for all values of \( x \)), we need to ensure that the coefficients of \( x^2 \), \( x \), and the constant term are all equal to zero.

### Step 1: Set the coefficient of \( x^2 \) to zero

The coefficient of \( x^2 \) is given by:

\[ (p-1)(2p+1) \]

To make the equation an identity, we set this coefficient to zero:

\[ (p-1)(2p+1) = 0 \]

This gives us two cases:

1. \( p - 1 = 0 \) which implies \( p = 1 \)
2. \( 2p + 1 = 0 \) which implies \( p = -\frac{1}{2} \)

### Step 2: Set the coefficient of \( x \) to zero

The coefficient of \( x \) is:

\[ p^2 - 1 \]

Setting this equal to zero gives:

\[ p^2 - 1 = 0 \]

Factoring this, we have:

\[ (p - 1)(p + 1) = 0 \]

This results in:

1. \( p - 1 = 0 \) which implies \( p = 1 \)
2. \( p + 1 = 0 \) which implies \( p = -1 \)

### Step 3: Set the constant term to zero

The constant term is:

\[ (p-1)(p-3) \]

Setting this equal to zero gives:

\[ (p-1)(p-3) = 0 \]

This results in:

1. \( p - 1 = 0 \) which implies \( p = 1 \)
2. \( p - 3 = 0 \) which implies \( p = 3 \)

### Step 4: Find the common value of \( p \)

Now, we have the following possible values for \( p \) from each step:

- From the \( x^2 \) coefficient: \( p = 1 \) or \( p = -\frac{1}{2} \)
- From the \( x \) coefficient: \( p = 1 \) or \( p = -1 \)
- From the constant term: \( p = 1 \) or \( p = 3 \)

The only common value across all three conditions is:

\[ p = 1 \]

### Conclusion

Thus, the value of \( p \) for which the given equation is an identity is:

\[ \boxed{1} \]