If `(x-1) and (x-2)` are two factors of `x^(3)-ax^(2)+14x +b` then find the value of a and b.

Answer

To find the values of \( a \) and \( b \) in the polynomial \( x^3 - ax^2 + 14x + b \) given that \( (x-1) \) and \( (x-2) \) are factors, we can follow these steps:

### Step 1: Use the Factor Theorem
Since \( (x-1) \) is a factor, substituting \( x = 1 \) into the polynomial should yield 0:
\[ 1^3 - a(1^2) + 14(1) + b = 0 \]
This simplifies to:
\[ 1 - a + 14 + b = 0 \]
Combining like terms gives us:
\[ -a + b + 15 = 0 \]
Rearranging this, we get our first equation:
\[ a - b = 15 \quad \text{(Equation 1)} \]

### Step 2: Substitute for the Second Factor
Next, since \( (x-2) \) is also a factor, substituting \( x = 2 \) into the polynomial should also yield 0:
\[ 2^3 - a(2^2) + 14(2) + b = 0 \]
This simplifies to:
\[ 8 - 4a + 28 + b = 0 \]
Combining like terms gives us:
\[ -4a + b + 36 = 0 \]
Rearranging this, we get our second equation:
\[ 4a - b = 36 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have a system of two equations:
1. \( a - b = 15 \)
2. \( 4a - b = 36 \)

We can solve these equations by elimination. Subtract Equation 1 from Equation 2:
\[ (4a - b) - (a - b) = 36 - 15 \]
This simplifies to:
\[ 4a - b - a + b = 21 \]
Thus,
\[ 3a = 21 \]
Dividing both sides by 3 gives:
\[ a = 7 \]

### Step 4: Substitute \( a \) Back to Find \( b \)
Now that we have \( a \), we can substitute it back into Equation 1 to find \( b \):
\[ 7 - b = 15 \]
Rearranging gives:
\[ -b = 15 - 7 \]
Thus,
\[ -b = 8 \quad \Rightarrow \quad b = -8 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ a = 7, \quad b = -8 \]

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