For what value of k, HCF of `2x^(2)+kx -12 and x^(2) +x-2k -2` is `(x+4)`.

Options

A. 5
B. 7
C. 10
D. `-4`

Answer

To find the value of \( k \) such that the highest common factor (HCF) of the polynomials \( 2x^2 + kx - 12 \) and \( x^2 + x - 2k - 2 \) is \( (x + 4) \), we can follow these steps:

### Step 1: Set the condition for HCF
Since \( (x + 4) \) is the HCF, it means that when we substitute \( x = -4 \) into both polynomials, the result should be zero.

### Step 2: Substitute \( x = -4 \) into the first polynomial
We substitute \( x = -4 \) into the first polynomial \( 2x^2 + kx - 12 \):

\[ 2(-4)^2 + k(-4) - 12 = 0 \]

Calculating this gives:

\[ 2(16) - 4k - 12 = 0 \]

\[ 32 - 4k - 12 = 0 \]

\[ 20 - 4k = 0 \]

\[ 4k = 20 \]

\[ k = 5 \]

### Step 3: Substitute \( x = -4 \) into the second polynomial
Now we substitute \( x = -4 \) into the second polynomial \( x^2 + x - 2k - 2 \):

\[ (-4)^2 + (-4) - 2k - 2 = 0 \]

Calculating this gives:

\[ 16 - 4 - 2k - 2 = 0 \]

\[ 10 - 2k = 0 \]

\[ 2k = 10 \]

\[ k = 5 \]

### Conclusion
Thus, the value of \( k \) for which the HCF of the two polynomials is \( (x + 4) \) is:

\[ \boxed{5} \]