When `f(x)=x^(3)+ax^(2)-bx-8` is divided by x-2, the remainder is zero and when divided by x+1, the remainder is -30. Find the values of 'a' and 'b'.

Answer

To solve the problem, we need to find the values of 'a' and 'b' in the polynomial \( f(x) = x^3 + ax^2 - bx - 8 \) given the conditions of the remainders when divided by \( x - 2 \) and \( x + 1 \).

### Step 1: Use the Remainder Theorem for \( x - 2 \)

According to the Remainder Theorem, if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). Since the remainder when \( f(x) \) is divided by \( x - 2 \) is 0, we have:

\[ f(2) = 0 \]

Substituting \( x = 2 \) into \( f(x) \):

\[ f(2) = 2^3 + a(2^2) - b(2) - 8 \]

Calculating this gives:

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

Setting this equal to 0:

\[ 4a - 2b = 0 \]

Dividing the entire equation by 2:

\[ 2a - b = 0 \quad \text{(Equation 1)} \]

### Step 2: Use the Remainder Theorem for \( x + 1 \)

Now, we know that when \( f(x) \) is divided by \( x + 1 \), the remainder is -30. Thus:

\[ f(-1) = -30 \]

Substituting \( x = -1 \) into \( f(x) \):

\[ f(-1) = (-1)^3 + a(-1)^2 - b(-1) - 8 \]

Calculating this gives:

\[ f(-1) = -1 + a + b - 8 = a + b - 9 \]

Setting this equal to -30:

\[ a + b - 9 = -30 \]

Adding 9 to both sides:

\[ a + b = -21 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

Now we have a system of two equations:

1. \( 2a - b = 0 \) (Equation 1)
2. \( a + b = -21 \) (Equation 2)

From Equation 1, we can express \( b \) in terms of \( a \):

\[ b = 2a \]

Now, substitute \( b = 2a \) into Equation 2:

\[ a + 2a = -21 \]

This simplifies to:

\[ 3a = -21 \]

Dividing both sides by 3:

\[ a = -7 \]

### Step 4: Find the Value of \( b \)

Now substitute \( a = -7 \) back into the equation \( b = 2a \):

\[ b = 2(-7) = -14 \]

### Conclusion

Thus, the values of \( a \) and \( b \) are:

\[ a = -7, \quad b = -14 \]