If 1,-2 and 3 are roots of
`x^3-2x^2+ax+6=0` , then find a .

Answer

To find the value of \( a \) in the polynomial equation \( x^3 - 2x^2 + ax + 6 = 0 \) given that the roots are \( 1, -2, \) and \( 3 \), we can follow these steps:

### Step 1: Use Vieta's Formulas
According to Vieta's formulas, for a cubic polynomial \( x^3 + bx^2 + cx + d = 0 \) with roots \( r_1, r_2, r_3 \):
- The sum of the roots \( r_1 + r_2 + r_3 = -b \)
- The sum of the product of the roots taken two at a time \( r_1r_2 + r_2r_3 + r_3r_1 = c \)
- The product of the roots \( r_1r_2r_3 = -d \)

### Step 2: Identify the Coefficients
From the given polynomial \( x^3 - 2x^2 + ax + 6 = 0 \), we can identify:
- \( b = -2 \)
- \( c = a \)
- \( d = 6 \)

### Step 3: Calculate the Sum of the Roots
The roots are \( 1, -2, \) and \( 3 \):
\[ 1 + (-2) + 3 = 1 + 3 - 2 = 2 \]
According to Vieta's, this should equal \( -(-2) = 2 \), which is consistent.

### Step 4: Calculate the Sum of the Product of the Roots Taken Two at a Time
Now, we calculate:
\[ (1)(-2) + (-2)(3) + (3)(1) = -2 - 6 + 3 = -5 \]
According to Vieta's, this should equal \( a \). Therefore, we have:
\[ a = -5 \]

### Step 5: Conclusion
Thus, the value of \( a \) is \( -5 \).

### Final Answer
\[ \boxed{-5} \]

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