For what value of a, the HCF of `y^(2) - 2y - 24` and `y^(2) - ay - 6` is (y-6) ?

Options

A. 15
B. 5
C. 30
D. 6

Answer

To find the value of \( a \) such that the HCF of the polynomials \( y^2 - 2y - 24 \) and \( y^2 - ay - 6 \) is \( (y - 6) \), we can follow these steps:

### Step 1: Factor the first polynomial
We start with the polynomial \( y^2 - 2y - 24 \). We need to factor it.

\[ y^2 - 2y - 24 = (y - 6)(y + 4) \]

### Step 2: Substitute \( y = 6 \)
Since we know that \( y - 6 \) is a factor, we can substitute \( y = 6 \) into the second polynomial \( y^2 - ay - 6 \) to find the value of \( a \).

\[ y^2 - ay - 6 \Rightarrow 6^2 - a(6) - 6 \]

Calculating this gives:

\[ 36 - 6a - 6 = 0 \]

### Step 3: Simplify the equation
Now, we simplify the equation:

\[ 36 - 6 - 6a = 0 \]

This simplifies to:

\[ 30 - 6a = 0 \]

### Step 4: Solve for \( a \)
Now, we can solve for \( a \):

\[ 30 = 6a \]

Dividing both sides by 6 gives:

\[ a = \frac{30}{6} = 5 \]

### Conclusion
Thus, the value of \( a \) for which the HCF of the given polynomials is \( (y - 6) \) is:

\[ \boxed{5} \]