Find the least integral value of 'k' for which the quadratic polynomial
`(k-1)x^(2) + 8x + k + 5 gt 0 AA x in R`

Answer

To find the least integral value of \( k \) for which the quadratic polynomial \( (k-1)x^2 + 8x + (k+5) > 0 \) for all \( x \in \mathbb{R} \), we need to ensure that the quadratic has no real roots and opens upwards. This can be determined by checking the following conditions:

1. The coefficient of \( x^2 \) must be positive.
2. The discriminant of the quadratic must be less than zero.

### Step 1: Ensure the coefficient of \( x^2 \) is positive
The coefficient of \( x^2 \) is \( k - 1 \). For the quadratic to open upwards, we need:
\[ k - 1 > 0 \implies k > 1 \]

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = k - 1 \)
- \( b = 8 \)
- \( c = k + 5 \)

Thus, the discriminant becomes:
\[ D = 8^2 - 4(k - 1)(k + 5) \]
Calculating this gives:
\[ D = 64 - 4[(k - 1)(k + 5)] \]
Expanding the product:
\[ (k - 1)(k + 5) = k^2 + 5k - k - 5 = k^2 + 4k - 5 \]
Substituting back into the discriminant:
\[ D = 64 - 4(k^2 + 4k - 5) = 64 - 4k^2 - 16k + 20 \]
Simplifying this:
\[ D = 84 - 4k^2 - 16k \]

### Step 3: Set the discriminant less than zero
To ensure the quadratic has no real roots, we need:
\[ 84 - 4k^2 - 16k < 0 \]
Rearranging gives:
\[ 4k^2 + 16k - 84 > 0 \]
Dividing the entire inequality by 4:
\[ k^2 + 4k - 21 > 0 \]

### Step 4: Factor the quadratic inequality
We can factor \( k^2 + 4k - 21 \):
\[ k^2 + 4k - 21 = (k + 7)(k - 3) \]
Now we need to find the intervals where this product is greater than zero. The roots are \( k = -7 \) and \( k = 3 \).

### Step 5: Test intervals
The critical points divide the number line into intervals:
1. \( (-\infty, -7) \)
2. \( (-7, 3) \)
3. \( (3, \infty) \)

We test a point from each interval:
- For \( k = -8 \) (in \( (-\infty, -7) \)): \( (-8 + 7)(-8 - 3) = (-1)(-11) > 0 \)
- For \( k = 0 \) (in \( (-7, 3) \)): \( (0 + 7)(0 - 3) = (7)(-3) < 0 \)
- For \( k = 4 \) (in \( (3, \infty) \)): \( (4 + 7)(4 - 3) = (11)(1) > 0 \)

Thus, the solution to the inequality \( (k + 7)(k - 3) > 0 \) is:
\[ k < -7 \quad \text{or} \quad k > 3 \]

### Step 6: Combine with the first condition
Since we also have \( k > 1 \), we combine this with \( k > 3 \). Therefore, the least integral value of \( k \) that satisfies both conditions is:
\[ k = 4 \]

### Final Answer
The least integral value of \( k \) for which the quadratic polynomial is greater than zero for all \( x \in \mathbb{R} \) is:
\[ \boxed{4} \]