x+2 is a factor of polynomial `ax^(3)+bx^(2)+x-2` and the remainder 4 is obtained on dividing this polynomial by (x-2). Find the value of a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given that \( x + 2 \) is a factor of the polynomial \( ax^3 + bx^2 + x - 2 \) and the remainder is 4 when this polynomial is divided by \( x - 2 \).

### Step 1: Use the Factor Theorem
Since \( x + 2 \) is a factor of the polynomial, we can use the Factor Theorem, which states that if \( x - c \) is a factor of a polynomial \( P(x) \), then \( P(c) = 0 \).

Here, we set \( x + 2 = 0 \) which gives us \( x = -2 \). Thus, we need to evaluate \( P(-2) \) and set it equal to 0.

### Step 2: Evaluate \( P(-2) \)
Substituting \( x = -2 \) into the polynomial:

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

Calculating this gives:

\[ P(-2) = a(-8) + b(4) - 2 - 2 = -8a + 4b - 4 \]

Setting this equal to 0 (since \( P(-2) = 0 \)):

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

### Step 3: Simplify the Equation
Rearranging the equation gives:

\[ -8a + 4b = 4 \]

Dividing the entire equation by 4:

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

### Step 4: Use the Remainder Theorem
Now, we know that the remainder when dividing by \( x - 2 \) is 4. According to the Remainder Theorem, \( P(2) = 4 \).

### Step 5: Evaluate \( P(2) \)
Substituting \( x = 2 \) into the polynomial:

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

Calculating this gives:

\[ P(2) = a(8) + b(4) + 2 - 2 = 8a + 4b \]

Setting this equal to 4 (since \( P(2) = 4 \)):

\[ 8a + 4b = 4 \]

### Step 6: Simplify the Second Equation
Dividing the entire equation by 4:

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

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

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

We can solve these equations simultaneously.

### Step 8: Subtract Equation 1 from Equation 2
Subtract Equation 1 from Equation 2:

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

This simplifies to:

\[ 4a = 0 \]

Thus,

\[ a = 0 \]

### Step 9: Substitute \( a \) Back to Find \( b \)
Now substitute \( a = 0 \) back into either equation. Using Equation 2:

\[ 2(0) + b = 1 \]

This gives:

\[ b = 1 \]

### Final Answer
The values of \( a \) and \( b \) are:

\[ a = 0, \quad b = 1 \]