What is the HCF of `(x^(2) +bx -x -b) and [x^(2) +x (a-1)-a]` ?

Options

A. `x+b`
B. `x+a`
C. `x+1`
D. `x-1 `

Answer

To find the Highest Common Factor (HCF) of the polynomials \( f(x) = x^2 + bx - x - b \) and \( g(x) = x^2 + x(a - 1) - a \), we will follow these steps:

### Step 1: Simplify \( f(x) \)

Start with the polynomial \( f(x) \):
\[ f(x) = x^2 + bx - x - b \]
We can rearrange and group the terms:
\[ f(x) = x^2 + (b - 1)x - b \]
Now, we will factor \( f(x) \). We can factor out \( x \) from the first two terms:
\[ f(x) = x(x + (b - 1)) - b \]
This doesn't lead us to a clear factorization. Instead, let's try to factor it directly:
\[ f(x) = x^2 + (b - 1)x - b \]
To factor it, we look for two numbers that multiply to \(-b\) and add to \(b - 1\).

### Step 2: Factor \( g(x) \)

Now, let's simplify \( g(x) \):
\[ g(x) = x^2 + x(a - 1) - a \]
Rearranging gives:
\[ g(x) = x^2 + (a - 1)x - a \]
We can factor out \( x \) from the first two terms:
\[ g(x) = x(x + (a - 1)) - a \]
Again, we will look for two numbers that multiply to \(-a\) and add to \(a - 1\).

### Step 3: Factor both polynomials

After analyzing both polynomials, we can factor them as follows:
1. For \( f(x) \):
\[ f(x) = (x - 1)(x + b) \]
2. For \( g(x) \):
\[ g(x) = (x - 1)(x + a) \]

### Step 4: Identify the HCF

Now we can see that both \( f(x) \) and \( g(x) \) have a common factor:
- \( f(x) = (x - 1)(x + b) \)
- \( g(x) = (x - 1)(x + a) \)

The common factor is \( (x - 1) \).

### Conclusion

Thus, the HCF of \( f(x) \) and \( g(x) \) is:
\[ \text{HCF} = x - 1 \]