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 solve the problem, we need to find the values of \(a\) and \(b\) such that the polynomial \(f(x) = 2x^3 + ax^2 + bx - 6\) has \(x - 1\) as a factor and leaves a remainder of 2 when divided by \(x - 2\).

### Step 1: Use the Factor Theorem
Since \(x - 1\) is a factor of the polynomial, by 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 equation:
\[ a + b - 4 = 0 \quad \text{(1)} \]
or
\[ a + b = 4 \quad \text{(1)} \]

### Step 2: Use the Remainder Theorem
Next, since the polynomial leaves a remainder of 2 when divided by \(x - 2\), we have:
\[ f(2) = 2 \]
Substituting \(x = 2\) into the polynomial:
\[ f(2) = 2(2)^3 + a(2)^2 + b(2) - 6 = 2 \]
This simplifies to:
\[ 2(8) + 4a + 2b - 6 = 2 \]
\[ 16 + 4a + 2b - 6 = 2 \]
Combining like terms gives:
\[ 4a + 2b + 10 = 2 \]
Thus, we can write the equation:
\[ 4a + 2b = -8 \quad \text{(2)} \]

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

From equation (1), we can express \(b\) in terms of \(a\):
\[ b = 4 - a \quad \text{(3)} \]

Substituting equation (3) 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 substitute \(a = -8\) back into equation (3):
\[ b = 4 - (-8) = 4 + 8 = 12 \]

### Final Values
Thus, the values of \(a\) and \(b\) are:
\[ a = -8, \quad b = 12 \]

### Conclusion
The values of \(a\) and \(b\) respectively are \(-8\) and \(12\).