If `2x^(3) + ax^(2) + bx- 6` has `(x - 1)` as a factor and leaves a remainder 2 when divided by `(x - 2)`, then the values of a and b respectively are____

Options

A. 6, 8
B. `-6, 10`
C. `4, -5`
D. `-8, 12`

Answer

To find the values of \( a \) and \( b \) in the polynomial \( 2x^3 + ax^2 + bx - 6 \), given that \( (x - 1) \) is a factor and the polynomial leaves a remainder of 2 when divided by \( (x - 2) \), we can follow these steps:

### Step 1: Use the Factor Theorem
Since \( (x - 1) \) is a factor, according to the Factor Theorem, we have:
\[ f(1) = 0 \]
Substituting \( x = 1 \) into the polynomial:
\[ f(1) = 2(1)^3 + a(1)^2 + b(1) - 6 = 0 \]
This simplifies to:
\[ 2 + a + b - 6 = 0 \]
Thus, we can write the first equation:
\[ a + b - 4 = 0 \quad \text{(Equation 1)} \]

### Step 2: Use the Remainder Theorem
The polynomial leaves a remainder of 2 when divided by \( (x - 2) \), which means:
\[ f(2) = 2 \]
Substituting \( x = 2 \) into the polynomial:
\[ f(2) = 2(2)^3 + a(2)^2 + b(2) - 6 = 2 \]
This simplifies to:
\[ 16 + 4a + 2b - 6 = 2 \]
Thus, we can write the second equation:
\[ 4a + 2b + 10 = 2 \]
Rearranging gives us:
\[ 4a + 2b = -8 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have the system of equations:
1. \( a + b = 4 \)
2. \( 4a + 2b = -8 \)

From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = 4 - a \]

Substituting \( b \) into Equation 2:
\[ 4a + 2(4 - a) = -8 \]
Expanding this gives:
\[ 4a + 8 - 2a = -8 \]
Combining like terms results in:
\[ 2a + 8 = -8 \]
Subtracting 8 from both sides:
\[ 2a = -16 \]
Dividing by 2:
\[ a = -8 \]

### Step 4: Find \( b \)
Now substituting \( a = -8 \) back into Equation 1:
\[ -8 + b = 4 \]
So:
\[ b = 4 + 8 = 12 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ \boxed{-8} \text{ and } \boxed{12} \]