If x + 1 is a factor of `x^(4) + (p - 3) x^(3) - (3p - 5) x^(2) + (2p - 9) x + 12` , then value of p is

Options

A. `-2`
B. `5`
C. `4`
D. `-1`

Answer

To find the value of \( p \) such that \( x + 1 \) is a factor of the polynomial

\[ f(x) = x^4 + (p - 3)x^3 - (3p - 5)x^2 + (2p - 9)x + 12, \]

we can use the Remainder Theorem. According to the Remainder Theorem, if \( x + 1 \) is a factor of \( f(x) \), then \( f(-1) = 0 \).

### Step 1: Substitute \( x = -1 \) into \( f(x) \)

We need to calculate \( f(-1) \):

\[ f(-1) = (-1)^4 + (p - 3)(-1)^3 - (3p - 5)(-1)^2 + (2p - 9)(-1) + 12. \]

### Step 2: Simplify the expression

Calculating each term:

- \( (-1)^4 = 1 \)
- \( (p - 3)(-1)^3 = -(p - 3) = -p + 3 \)
- \( -(3p - 5)(-1)^2 = -(3p - 5) = -3p + 5 \)
- \( (2p - 9)(-1) = -2p + 9 \)
- The constant term is \( 12 \)

Now, substituting these values into the equation:

\[ f(-1) = 1 - p + 3 - 3p + 5 - 2p + 9 + 12. \]

### Step 3: Combine like terms

Combining all the terms:

\[ f(-1) = 1 + 3 + 5 + 9 + 12 - p - 3p - 2p. \]

This simplifies to:

\[ f(-1) = 30 - 6p. \]

### Step 4: Set the equation to zero

Since \( f(-1) = 0 \):

\[ 30 - 6p = 0. \]

### Step 5: Solve for \( p \)

Rearranging the equation gives:

\[ 6p = 30 \implies p = \frac{30}{6} = 5. \]

Thus, the value of \( p \) is

\[ \boxed{5}. \]