For what value of p, ( - 4) is a zero of the polynomial `x^(2) - 2x - ( 7p+3) ` ?

Answer

To find the value of \( p \) for which \( -4 \) is a zero of the polynomial \( x^2 - 2x - (7p + 3) \), we can follow these steps:

### Step 1: Substitute the zero into the polynomial
Since \( -4 \) is a zero of the polynomial, we can substitute \( x = -4 \) into the polynomial equation:

\[ (-4)^2 - 2(-4) - (7p + 3) = 0 \]

### Step 2: Calculate \( (-4)^2 \)
Calculating \( (-4)^2 \):

\[ 16 \]

### Step 3: Calculate \( -2(-4) \)
Calculating \( -2(-4) \):

\[ 8 \]

### Step 4: Substitute values into the equation
Now substituting the calculated values back into the equation:

\[ 16 + 8 - (7p + 3) = 0 \]

### Step 5: Simplify the equation
Now simplify the equation:

\[ 16 + 8 - 3 - 7p = 0 \]

This simplifies to:

\[ 21 - 7p = 0 \]

### Step 6: Solve for \( p \)
Now, isolate \( p \):

\[ 7p = 21 \]

Dividing both sides by 7 gives:

\[ p = 3 \]

### Final Answer
Thus, the value of \( p \) is:

\[ \boxed{3} \]

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