If (x - a) is the HCF of $x^{2} + x - 12 and 2x^{2} - ax - 9$, then the value of 'a' is :

Q: If (x - a) is the HCF of $x^{2} + x - 12 and 2x^{2} - ax - 9$, then the value of 'a' is :

To determine the value of 'a' such that $(x - a)$ is the HCF of the polynomials $x^2 + x - 12$ and $2x^2 - ax - 9$, we need to ensure that $(x - a)$ divides both polynomials. ### Step-by-Step Explanation: 1. **Factor the First Polynomial:** - The polynomial $x^2 + x - 12$ can be factored as $(x - 3)(x + 4)$. - Therefore, the roots of this polynomial are $x = 3$ and $x = -4$. 2. **Substitute the Roots into the Second Polynomial:** - Since $(x - a)$ is a common factor, it must also be a root of the second polynomial $2x^2 - ax - 9$. 3. **Check for $x = 3$:** - Substitute $x = 3$ into $2x^2 - ax - 9$: \[ 2(3)^2 - a(3) - 9 = 0 \\ 18 - 3a - 9 = 0 \\ 9 - 3a = 0 \\ 3a = 9 \\ a = 3 \] - This calculation shows that when $x = 3$, the polynomial $2x^2 - ax - 9$ equals zero if $a = 3$. 4. **Verify the Other Root $x = -4$:** - Substitute $x = -4$ into $2x^2 - ax - 9$: \[ 2(-4)^2 - a(-4) - 9 = 0 \\ 32 + 4a - 9 = 0 \\ 4a + 23 = 0 \\ 4a = -23 \] - This calculation does not yield an integer value for $a$, confirming that $x = -4$ is not a valid root for the second polynomial when $a = 3$. ### Conclusion: - **Option 1 (a = 3) is Correct:** - The value $a = 3$ satisfies the condition that $(x - a)$ is a factor of both polynomials, as shown by the substitution of $x = 3$ into the second polynomial. - **Why Other Options are Incorrect:** - **Option 2 (-3):** Substituting $x = -3$ into the second polynomial does not yield zero, hence it is not a factor. - **Option 3 (-4):** As shown, substituting $x = -4$ does not satisfy the condition for $a$ to be an integer. - **Option 4 (4):** Substituting $x = 4$ into the second polynomial does not yield zero, hence it is not a factor. Thus, the correct answer is Option 1, where $a = 3$.

Question

Accepted Answer

To determine the value of 'a' such that $(x - a)$ is the HCF of the polynomials $x^2 + x - 12$ and $2x^2 - ax - 9$, we need to ensure that $(x - a)$ divides both polynomials.

### Step-by-Step Explanation:

1. **Factor the First Polynomial:**
   - The polynomial $x^2 + x - 12$ can be factored as $(x - 3)(x + 4)$.
   - Therefore, the roots of this polynomial are $x = 3$ and $x = -4$.

2. **Substitute the Roots into the Second Polynomial:**
   - Since $(x - a)$ is a common factor, it must also be a root of the second polynomial $2x^2 - ax - 9$.

3. **Check for $x = 3$:**
   - Substitute $x = 3$ into $2x^2 - ax - 9$:
     $$
     2(3)^2 - a(3) - 9 = 0 \
     18 - 3a - 9 = 0 \
     9 - 3a = 0 \
     3a = 9 \
     a = 3
     $$
   - This calculation shows that when $x = 3$, the polynomial $2x^2 - ax - 9$ equals zero if $a = 3$.

4. **Verify the Other Root $x = -4$:**
   - Substitute $x = -4$ into $2x^2 - ax - 9$:
     $$
     2(-4)^2 - a(-4) - 9 = 0 \
     32 + 4a - 9 = 0 \
     4a + 23 = 0 \
     4a = -23
     $$
   - This calculation does not yield an integer value for $a$, confirming that $x = -4$ is not a valid root for the second polynomial when $a = 3$.

### Conclusion:

- **Option 1 (a = 3) is Correct:**
  - The value $a = 3$ satisfies the condition that $(x - a)$ is a factor of both polynomials, as shown by the substitution of $x = 3$ into the second polynomial.

- **Why Other Options are Incorrect:**
  - **Option 2 (-3):** Substituting $x = -3$ into the second polynomial does not yield zero, hence it is not a factor.
  - **Option 3 (-4):** As shown, substituting $x = -4$ does not satisfy the condition for $a$ to be an integer.
  - **Option 4 (4):** Substituting $x = 4$ into the second polynomial does not yield zero, hence it is not a factor.

Thus, the correct answer is Option 1, where $a = 3$.