Find the values of constants a and b when x- 2 and x + 3 both are the factors of expression `x^3+ax^2+bx -12.`

Answer

To solve the problem of finding the constants \( a \) and \( b \) in the expression \( x^3 + ax^2 + bx - 12 \) given that \( x - 2 \) and \( x + 3 \) are factors, we can follow these steps:

### Step 1: Use the Factor Theorem
Since \( x - 2 \) is a factor, substituting \( x = 2 \) in the expression should yield 0.

\[ 2^3 + a(2^2) + b(2) - 12 = 0 \]

### Step 2: Simplify the Equation
Calculating \( 2^3 \) and \( 2^2 \):

\[ 8 + 4a + 2b - 12 = 0 \]

This simplifies to:

\[ 4a + 2b - 4 = 0 \]

### Step 3: Rearrange the Equation
Rearranging gives us:

\[ 4a + 2b = 4 \]

Dividing the entire equation by 2:

\[ 2a + b = 2 \quad \text{(Equation 1)} \]

### Step 4: Use the Second Factor
Now, since \( x + 3 \) is also a factor, substituting \( x = -3 \) in the expression should also yield 0.

\[ (-3)^3 + a(-3)^2 + b(-3) - 12 = 0 \]

### Step 5: Simplify the Second Equation
Calculating \( (-3)^3 \) and \( (-3)^2 \):

\[ -27 + 9a - 3b - 12 = 0 \]

This simplifies to:

\[ 9a - 3b - 39 = 0 \]

### Step 6: Rearrange the Second Equation
Rearranging gives us:

\[ 9a - 3b = 39 \]

Dividing the entire equation by 3:

\[ 3a - b = 13 \quad \text{(Equation 2)} \]

### Step 7: Solve the System of Equations
Now we have a system of equations:

1. \( 2a + b = 2 \) (Equation 1)
2. \( 3a - b = 13 \) (Equation 2)

Adding both equations to eliminate \( b \):

\[ (2a + b) + (3a - b) = 2 + 13 \]

This simplifies to:

\[ 5a = 15 \]

### Step 8: Find \( a \)
Solving for \( a \):

\[ a = \frac{15}{5} = 3 \]

### Step 9: Substitute \( a \) Back to Find \( b \)
Substituting \( a = 3 \) back into Equation 1:

\[ 2(3) + b = 2 \]

This simplifies to:

\[ 6 + b = 2 \]

### Step 10: Solve for \( b \)
Solving for \( b \):

\[ b = 2 - 6 = -4 \]

### Final Answer
Thus, the values of the constants are:

\[ a = 3, \quad b = -4 \]

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