If `2x^(2) + ax + 2b`, when divided by `x-1`, leaves a remainder of 16 and `x^(2) + bx + 2a`, when divided by `x+1`, leaves a remainder of -1. then a + b equals:
(a)14
(b) -8
(c) -14
(d)8

Options

A. 14
B. `-8`
C. `-14`
D. 8

Answer

To solve the problem step by step, we will use the Remainder Theorem, which states that the remainder of the division of a polynomial \( f(x) \) by \( x - c \) is equal to \( f(c) \).

### Step 1: Set up the first equation
We are given that when \( 2x^2 + ax + 2b \) is divided by \( x - 1 \), the remainder is 16. According to the Remainder Theorem, we can substitute \( x = 1 \) into the polynomial:

\[ f(1) = 2(1)^2 + a(1) + 2b = 16 \]

This simplifies to:

\[ 2 + a + 2b = 16 \]

Rearranging gives us:

\[ a + 2b = 14 \quad \text{(Equation 1)} \]

### Step 2: Set up the second equation
Next, we are told that when \( x^2 + bx + 2a \) is divided by \( x + 1 \), the remainder is -1. Again, using the Remainder Theorem, we substitute \( x = -1 \):

\[ f(-1) = (-1)^2 + b(-1) + 2a = -1 \]

This simplifies to:

\[ 1 - b + 2a = -1 \]

Rearranging gives us:

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

### Step 3: Solve the system of equations
Now we have a system of two equations:

1. \( a + 2b = 14 \)
2. \( 2a - b = -2 \)

We can solve these equations simultaneously. Let's express \( b \) from Equation 1:

\[ b = \frac{14 - a}{2} \]

Now, substitute this expression for \( b \) into Equation 2:

\[ 2a - \left(\frac{14 - a}{2}\right) = -2 \]

Multiply through by 2 to eliminate the fraction:

\[ 4a - (14 - a) = -4 \]

This simplifies to:

\[ 4a - 14 + a = -4 \]

Combining like terms gives:

\[ 5a - 14 = -4 \]

Adding 14 to both sides:

\[ 5a = 10 \]

Dividing by 5:

\[ a = 2 \]

### Step 4: Find \( b \)
Now that we have \( a \), we can substitute it back into Equation 1 to find \( b \):

\[ 2 + 2b = 14 \]

Subtracting 2 from both sides:

\[ 2b = 12 \]

Dividing by 2:

\[ b = 6 \]

### Step 5: Calculate \( a + b \)
Finally, we need to find \( a + b \):

\[ a + b = 2 + 6 = 8 \]

Thus, the answer is:

\[ \boxed{8} \]