Suppose 1,2,3 are the roots of the equation `x^4 + ax^2 + bx + c = 0`. Find the value of c.

Answer

To find the value of \( c \) in the equation \( x^4 + ax^2 + bx + c = 0 \) given that the roots are \( 1, 2, \) and \( 3 \), we can follow these steps:

### Step 1: Set up the equations using the roots
Since \( 1, 2, \) and \( 3 \) are roots of the polynomial, we can substitute these values into the polynomial equation to create a system of equations.

1. **For \( x = 1 \)**:
\[ 1^4 + a(1^2) + b(1) + c = 0 \implies 1 + a + b + c = 0 \implies a + b + c = -1 \quad \text{(Equation 1)} \]

2. **For \( x = 2 \)**:
\[ 2^4 + a(2^2) + b(2) + c = 0 \implies 16 + 4a + 2b + c = 0 \implies 4a + 2b + c = -16 \quad \text{(Equation 2)} \]

3. **For \( x = 3 \)**:
\[ 3^4 + a(3^2) + b(3) + c = 0 \implies 81 + 9a + 3b + c = 0 \implies 9a + 3b + c = -81 \quad \text{(Equation 3)} \]

### Step 2: Solve the system of equations
Now we have the following system of equations:
1. \( a + b + c = -1 \) (Equation 1)
2. \( 4a + 2b + c = -16 \) (Equation 2)
3. \( 9a + 3b + c = -81 \) (Equation 3)

#### Subtract Equation 1 from Equation 2:
\[ (4a + 2b + c) - (a + b + c) = -16 - (-1) \]
This simplifies to:
\[ 3a + b = -15 \quad \text{(Equation 4)} \]

#### Subtract Equation 1 from Equation 3:
\[ (9a + 3b + c) - (a + b + c) = -81 - (-1) \]
This simplifies to:
\[ 8a + 2b = -80 \quad \text{(Equation 5)} \]

### Step 3: Solve Equations 4 and 5
Now we can solve Equations 4 and 5:
1. From Equation 4: \( b = -15 - 3a \)
2. Substitute \( b \) in Equation 5:
\[ 8a + 2(-15 - 3a) = -80 \]
Simplifying gives:
\[ 8a - 30 - 6a = -80 \implies 2a - 30 = -80 \implies 2a = -50 \implies a = -25 \]

### Step 4: Find \( b \)
Substituting \( a = -25 \) back into Equation 4:
\[ 3(-25) + b = -15 \implies -75 + b = -15 \implies b = 60 \]

### Step 5: Find \( c \)
Now substitute \( a \) and \( b \) back into Equation 1:
\[ -25 + 60 + c = -1 \implies 35 + c = -1 \implies c = -1 - 35 = -36 \]

### Conclusion
The value of \( c \) is:
\[ \boxed{-36} \]