If 1,2,3 are the roots of the equation ` x^(3) + ax^(2) + bx + c=0` , then

Options

A. a=1, b=2, c=3
B. a= -6,b =11 , c= - 6
C. a = 6, b=11, c= 6
D. a=6, b= 6 , c= 6

Answer

To find the values of \( a \), \( b \), and \( c \) in the polynomial equation \( x^3 + ax^2 + bx + c = 0 \) given that the roots are 1, 2, and 3, we can use Vieta's formulas and the fact that the roots satisfy the equation.

### Step-by-step Solution:

1. **Using the roots in the polynomial:**
Since 1, 2, and 3 are the roots of the polynomial, we can substitute these values into the equation to form a system of equations.

- For \( x = 1 \):
\[ 1^3 + a(1^2) + b(1) + c = 0 \implies 1 + a + b + c = 0 \tag{1} \]

- For \( x = 2 \):
\[ 2^3 + a(2^2) + b(2) + c = 0 \implies 8 + 4a + 2b + c = 0 \tag{2} \]

- For \( x = 3 \):
\[ 3^3 + a(3^2) + b(3) + c = 0 \implies 27 + 9a + 3b + c = 0 \tag{3} \]

2. **Expressing \( c \) from Equation (1):**
From Equation (1):
\[ c = -1 - a - b \tag{4} \]

3. **Substituting \( c \) into Equations (2) and (3):**
Substitute Equation (4) into Equation (2):
\[ 8 + 4a + 2b + (-1 - a - b) = 0 \]
Simplifying this gives:
\[ 7 + 3a + b = 0 \implies 3a + b = -7 \tag{5} \]

Now substitute Equation (4) into Equation (3):
\[ 27 + 9a + 3b + (-1 - a - b) = 0 \]
Simplifying this gives:
\[ 26 + 8a + 2b = 0 \implies 8a + 2b = -26 \implies 4a + b = -13 \tag{6} \]

4. **Solving Equations (5) and (6):**
Now we have a system of two equations:
\[ 3a + b = -7 \tag{5} \]
\[ 4a + b = -13 \tag{6} \]

Subtract Equation (5) from Equation (6):
\[ (4a + b) - (3a + b) = -13 + 7 \]
This simplifies to:
\[ a = -6 \]

5. **Finding \( b \):**
Substitute \( a = -6 \) back into Equation (5):
\[ 3(-6) + b = -7 \implies -18 + b = -7 \implies b = 11 \]

6. **Finding \( c \):**
Substitute \( a = -6 \) and \( b = 11 \) back into Equation (4):
\[ c = -1 - (-6) - 11 = -1 + 6 - 11 = -6 \]

### Final Values:
Thus, the values are:
- \( a = -6 \)
- \( b = 11 \)
- \( c = -6 \)