If `f(x)=(x^(2)+3x+2)(x^(2)-7x+a)` and `g(x)=(x^(2)-x-12)(x^(2)+5x+b)`, then the value of `a` and `b`, if `(x+1)(x-4)` is H.C.F. of `f(x)` and `g(x)` is (a) a=10 : b=6` (b) `a=4 : b=12` (c)`a=12 : b=4` (d)`a=6 : b=10`

Options

A. `a=10 : b=6`
B. `a=4 : b=12`
C. `a=12 : b=4`
D. `a=6 : b=10`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) such that the given polynomials \( f(x) \) and \( g(x) \) have \( (x + 1)(x - 4) \) as their highest common factor (H.C.F.).

### Step 1: Factor \( f(x) \)
Given:
\[ f(x) = (x^2 + 3x + 2)(x^2 - 7x + a) \]
First, we factor \( x^2 + 3x + 2 \):
\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = (x + 1)(x + 2)(x^2 - 7x + a) \]

### Step 2: Determine conditions for \( f(x) \)
Since \( (x + 1)(x - 4) \) is a factor of \( f(x) \), we need \( x^2 - 7x + a \) to be zero when \( x = 4 \):
\[ 4^2 - 7(4) + a = 0 \]
Calculating:
\[ 16 - 28 + a = 0 \implies a - 12 = 0 \implies a = 12 \]

### Step 3: Factor \( g(x) \)
Given:
\[ g(x) = (x^2 - x - 12)(x^2 + 5x + b) \]
First, we factor \( x^2 - x - 12 \):
\[ x^2 - x - 12 = (x - 4)(x + 3) \]
Thus, we can rewrite \( g(x) \):
\[ g(x) = (x - 4)(x + 3)(x^2 + 5x + b) \]

### Step 4: Determine conditions for \( g(x) \)
Since \( (x + 1)(x - 4) \) is a factor of \( g(x) \), we need \( x^2 + 5x + b \) to be zero when \( x = -1 \):
\[ (-1)^2 + 5(-1) + b = 0 \]
Calculating:
\[ 1 - 5 + b = 0 \implies b - 4 = 0 \implies b = 4 \]

### Final Values
Thus, we have:
\[ a = 12 \quad \text{and} \quad b = 4 \]

### Conclusion
The correct option is \( (c) \, a = 12 : b = 4 \).

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