If `2x^(2) + ax +b` when divided by `x-3` leaves a remainder of 35, and `2x^(2) + bx +a` when divided by `x-3` leaves a remainder of 29, then what is the value of `a+b` ?
A)`-23`
B)7
C)`-7`
D)23

Options

A. `-23`
B. 7
C. `-7`
D. 23

Answer

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

### Step 1: Set up the equations using the Remainder Theorem
1. For the first polynomial \( 2x^2 + ax + b \):
- When divided by \( x - 3 \), the remainder is 35.
- Therefore, we can substitute \( x = 3 \) into the polynomial:
\[ f(3) = 2(3^2) + a(3) + b = 35 \]
- Simplifying this gives:
\[ 2(9) + 3a + b = 35 \implies 18 + 3a + b = 35 \]
- Rearranging this, we get:
\[ 3a + b = 17 \quad \text{(Equation 1)} \]

### Step 2: Set up the second equation
2. For the second polynomial \( 2x^2 + bx + a \):
- When divided by \( x - 3 \), the remainder is 29.
- Again, substituting \( x = 3 \):
\[ f(3) = 2(3^2) + b(3) + a = 29 \]
- Simplifying this gives:
\[ 2(9) + 3b + a = 29 \implies 18 + 3b + a = 29 \]
- Rearranging this, we get:
\[ 3b + a = 11 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have the following system of equations:
1. \( 3a + b = 17 \) (Equation 1)
2. \( 3b + a = 11 \) (Equation 2)

We can solve these equations simultaneously.

### Step 4: Express \( b \) in terms of \( a \)
From Equation 1:
\[ b = 17 - 3a \]

### Step 5: Substitute \( b \) into Equation 2
Substituting \( b \) into Equation 2:
\[ 3(17 - 3a) + a = 11 \]
Expanding this gives:
\[ 51 - 9a + a = 11 \]
Combining like terms:
\[ 51 - 8a = 11 \]
Rearranging gives:
\[ -8a = 11 - 51 \implies -8a = -40 \implies a = 5 \]

### Step 6: Find \( b \) using the value of \( a \)
Substituting \( a = 5 \) back into the expression for \( b \):
\[ b = 17 - 3(5) = 17 - 15 = 2 \]

### Step 7: Find \( a + b \)
Now we can find \( a + b \):
\[ a + b = 5 + 2 = 7 \]

### Final Answer
The value of \( a + b \) is \( 7 \).