Find the value of b for which the HCF of `x^(2) + 2bx + 3b + 3 and x^(2) +x- 5b` is `(x+5)`

Answer

To find the value of \( b \) for which the highest common factor (HCF) of the polynomials \( P(x) = x^2 + 2bx + 3b + 3 \) and \( Q(x) = x^2 + x - 5b \) is \( (x + 5) \), we can follow these steps:

### Step 1: Set up the condition for HCF
Since \( (x + 5) \) is a factor of both polynomials, we can substitute \( x = -5 \) into both polynomials and set them equal to zero.

### Step 2: Substitute \( x = -5 \) into \( Q(x) \)
Substituting \( x = -5 \) into \( Q(x) \):

\[ Q(-5) = (-5)^2 + (-5) - 5b \]
\[ = 25 - 5 - 5b \]
\[ = 20 - 5b \]

### Step 3: Set \( Q(-5) = 0 \)
Since \( (x + 5) \) is a factor, we set \( Q(-5) = 0 \):

\[ 20 - 5b = 0 \]

### Step 4: Solve for \( b \)
Rearranging the equation gives:

\[ 5b = 20 \]
\[ b = \frac{20}{5} = 4 \]

### Step 5: Verify by substituting \( x = -5 \) into \( P(x) \)
To ensure that \( (x + 5) \) is also a factor of \( P(x) \), we substitute \( x = -5 \) into \( P(x) \):

\[ P(-5) = (-5)^2 + 2b(-5) + 3b + 3 \]
Substituting \( b = 4 \):

\[ = 25 - 10(4) + 3(4) + 3 \]
\[ = 25 - 40 + 12 + 3 \]
\[ = 25 - 40 + 15 = 0 \]

Since \( P(-5) = 0 \), this confirms that \( (x + 5) \) is indeed a factor of \( P(x) \).

### Conclusion
Thus, the value of \( b \) for which the HCF of the two polynomials is \( (x + 5) \) is:

\[ \boxed{4} \]