If one of the zeroes of the cubic polynomial `x^(3)+ax^(2)+bx+c` is -1, then the perpendicular of the other two zeroes is

Options

A. `b-a+1`
B. `b-a-1`
C. `a-b+1`
D. `a-b-1`

Answer

To solve the problem step by step, we will analyze the cubic polynomial \( P(x) = x^3 + ax^2 + bx + c \) given that one of its zeroes is \( -1 \).

### Step 1: Substitute the zero into the polynomial
We know that \( -1 \) is a zero of the polynomial. Therefore, we can substitute \( x = -1 \) into the polynomial:

\[ P(-1) = (-1)^3 + a(-1)^2 + b(-1) + c = 0 \]

This simplifies to:

\[ -1 + a - b + c = 0 \]

Rearranging gives us:

\[ c = 1 - a + b \quad \text{(Equation 1)} \]

### Step 2: Use the relationship of the product of the zeroes
For a cubic polynomial, the product of the zeroes \( \alpha, \beta, \gamma \) (where \( \alpha = -1 \), \( \beta \), and \( \gamma \) are the zeroes) is given by:

\[ \alpha \cdot \beta \cdot \gamma = -\frac{c}{1} \]

Substituting \( \alpha = -1 \):

\[ -1 \cdot \beta \cdot \gamma = -c \]

This leads to:

\[ \beta \cdot \gamma = c \quad \text{(Equation 2)} \]

### Step 3: Substitute \( c \) from Equation 1 into Equation 2
From Equation 1, we have \( c = 1 - a + b \). Substituting this into Equation 2 gives:

\[ \beta \cdot \gamma = 1 - a + b \]

### Step 4: Identify the relationship between the zeroes
Now we can express the polynomial in terms of its factors. Since \( -1 \) is a zero, we can factor the polynomial as:

\[ P(x) = (x + 1)(x^2 + px + q) \]

where \( p \) and \( q \) are coefficients we need to determine. Expanding this gives:

\[ P(x) = x^3 + px^2 + qx + x^2 + px + q = x^3 + (p + 1)x^2 + (q + p)x + q \]

By comparing coefficients with \( P(x) = x^3 + ax^2 + bx + c \), we have:

1. \( p + 1 = a \)
2. \( q + p = b \)
3. \( q = c \)

### Step 5: Solve for \( p \) and \( q \)
From \( q = c \) and substituting \( c = 1 - a + b \):

\[ q = 1 - a + b \]

From \( p + 1 = a \), we can express \( p \) as:

\[ p = a - 1 \]

Substituting \( p \) into \( q + p = b \):

\[ (1 - a + b) + (a - 1) = b \]

This simplifies to:

\[ b = b \]

This confirms our relationships are consistent.

### Step 6: Find the perpendicular of the other two zeroes
The other two zeroes are \( \beta \) and \( \gamma \). The product of these zeroes is \( \beta \cdot \gamma = 1 - a + b \).

To find the perpendicular of the other two zeroes, we can apply the relationship of the sum of the zeroes:

\[ \beta + \gamma = -\frac{(p + 1)}{1} = -p - 1 = - (a - 1) - 1 = -a \]

Thus, the perpendicular of the other two zeroes can be calculated using the relationship of their sum and product.

### Final Result
The perpendicular of the other two zeroes is given by:

\[ \text{Perpendicular} = \sqrt{(\beta + \gamma)^2 - 4(\beta \cdot \gamma)} = \sqrt{(-a)^2 - 4(1 - a + b)} \]