If (1 - p) is a zero of the polynomial `x^2 +px + 1-p= 0`, then find both zeroes of the polynomial.

Options

A. `0,-1`
B. `1,-1`
C. 1,0
D. 0,0

Answer

To solve the problem, we need to find both zeroes of the polynomial \( x^2 + px + 1 - p = 0 \) given that \( (1 - p) \) is one of its zeroes.

### Step 1: Substitute the zero into the polynomial
Since \( (1 - p) \) is a zero of the polynomial, we can substitute \( x = 1 - p \) into the polynomial:

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

### Step 2: Expand the equation
Now, we will expand the left-hand side:

\[ (1 - p)^2 = 1 - 2p + p^2 \]
\[ p(1 - p) = p - p^2 \]
Substituting these into the equation gives:

\[ 1 - 2p + p^2 + p - p^2 + 1 - p = 0 \]

### Step 3: Simplify the equation
Now, combine like terms:

\[ 1 + 1 - 2p + p - p + p^2 - p^2 = 0 \]
This simplifies to:

\[ 2 - 2p = 0 \]

### Step 4: Solve for \( p \)
Now, we can solve for \( p \):

\[ 2 = 2p \implies p = 1 \]

### Step 5: Substitute \( p \) back into the polynomial
Now that we have \( p = 1 \), we substitute it back into the original polynomial:

\[ x^2 + 1x + 1 - 1 = 0 \]
This simplifies to:

\[ x^2 + x = 0 \]

### Step 6: Factor the polynomial
We can factor this polynomial:

\[ x(x + 1) = 0 \]

### Step 7: Find the zeroes
Setting each factor to zero gives us the zeroes:

1. \( x = 0 \)
2. \( x + 1 = 0 \implies x = -1 \)

### Conclusion
The two zeroes of the polynomial are \( 0 \) and \( -1 \).