Polynomial `x^3-ax^2+bx-6` leaves remainder -8 when divided by x-1 and x-2 is a factor of it. 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 \( f(x) = x^3 - ax^2 + bx - 6 \) given that it leaves a remainder of -8 when divided by \( x - 1 \) and that \( x - 2 \) is a factor of the polynomial.

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

According to the Remainder Theorem, if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). Here, we need to find \( f(1) \):

\[ f(1) = 1^3 - a(1^2) + b(1) - 6 \]
\[ = 1 - a + b - 6 \]
\[ = -a + b - 5 \]

Since the remainder is -8, we set up the equation:
\[ -a + b - 5 = -8 \]
\[ -a + b = -3 \quad \text{(Equation 1)} \]

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

Since \( x - 2 \) is a factor, we have \( f(2) = 0 \):

\[ f(2) = 2^3 - a(2^2) + b(2) - 6 \]
\[ = 8 - 4a + 2b - 6 \]
\[ = -4a + 2b + 2 \]

Setting this equal to 0 gives us:
\[ -4a + 2b + 2 = 0 \]
\[ -4a + 2b = -2 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

Now we have two equations:
1. \( -a + b = -3 \)
2. \( -4a + 2b = -2 \)

We can simplify Equation 2 by dividing everything by 2:
\[ -2a + b = -1 \quad \text{(Equation 3)} \]

Now we can solve Equations 1 and 3 together:
1. \( -a + b = -3 \)
2. \( -2a + b = -1 \)

Subtract Equation 1 from Equation 3:
\[ (-2a + b) - (-a + b) = -1 - (-3) \]
\[ -2a + b + a - b = 2 \]
\[ -a = 2 \quad \Rightarrow \quad a = -2 \]

### Step 4: Substitute \( a \) back to find \( b \)

Now substitute \( a = -2 \) back into Equation 1:
\[ -(-2) + b = -3 \]
\[ 2 + b = -3 \]
\[ b = -3 - 2 = -5 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:
\[ a = 2, \quad b = -1 \]

### Summary of the Solution Steps:
1. Set up the equation using the Remainder Theorem for \( x - 1 \).
2. Set up the equation using the Factor Theorem for \( x - 2 \).
3. Solve the system of equations obtained from steps 1 and 2.
4. Substitute back to find the values of \( a \) and \( b \).