If the roots of ` x^4 - 8x^3 + 14 x^2 + 8x - 15=0` are in A.P then the roots are

Options

A. `-1,1,3,5`
B. `1,1,-3,-5`
C. `1,3,5,7`
D. `1,2,3,4`

Answer

To find the roots of the polynomial equation \( x^4 - 8x^3 + 14x^2 + 8x - 15 = 0 \) given that the roots are in arithmetic progression (A.P.), we can follow these steps:

### Step 1: Assume the Roots
Since the roots are in A.P., we can denote them as:
- \( a - 3d \)
- \( a - d \)
- \( a + d \)
- \( a + 3d \)

### Step 2: Use the Sum of Roots
According to Vieta's formulas, the sum of the roots of the polynomial \( x^4 + bx^3 + cx^2 + dx + e = 0 \) is given by \( -\frac{b}{a} \). Here, \( b = -8 \) and \( a = 1 \), thus:
\[ \text{Sum of roots} = -\frac{-8}{1} = 8 \]
Setting up the equation for the sum of our assumed roots:
\[ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 8 \]
This simplifies to:
\[ 4a = 8 \implies a = 2 \]

### Step 3: Use the Product of Roots Taken Two at a Time
Next, we need to find the sum of the products of the roots taken two at a time, which is equal to the coefficient of \( x^2 \) (which is 14) in our polynomial:
\[ \text{Sum of products of roots taken two at a time} = 14 \]
Calculating this:
\[ (a - 3d)(a - d) + (a - 3d)(a + d) + (a - 3d)(a + 3d) + (a - d)(a + d) + (a - d)(a + 3d) + (a + d)(a + 3d) \]
Substituting \( a = 2 \):
\[ (2 - 3d)(2 - d) + (2 - 3d)(2 + d) + (2 - 3d)(2 + 3d) + (2 - d)(2 + d) + (2 - d)(2 + 3d) + (2 + d)(2 + 3d) = 14 \]

### Step 4: Simplify the Expression
Calculating each term:
1. \( (2 - 3d)(2 - d) = 4 - 2d - 6d + 3d^2 = 4 - 8d + 3d^2 \)
2. \( (2 - 3d)(2 + d) = 4 + 2d - 6d - 3d^2 = 4 - 4d - 3d^2 \)
3. \( (2 - 3d)(2 + 3d) = 4 + 6d - 6d - 9d^2 = 4 - 9d^2 \)
4. \( (2 - d)(2 + d) = 4 - d^2 \)
5. \( (2 - d)(2 + 3d) = 4 + 6d - 2d - 3d^2 = 4 + 4d - 3d^2 \)
6. \( (2 + d)(2 + 3d) = 4 + 6d + 2d + 3d^2 = 4 + 8d + 3d^2 \)

Summing these results gives:
\[ (4 - 8d + 3d^2) + (4 - 4d - 3d^2) + (4 - 9d^2) + (4 - d^2) + (4 + 4d - 3d^2) + (4 + 8d + 3d^2) \]
Combining like terms:
\[ 24 - 8d - 4d + 4d + 8d + (3d^2 - 3d^2 - 9d^2 - d^2 - 3d^2 + 3d^2) = 14 \]
This simplifies to:
\[ 24 - 10d^2 = 14 \]
### Step 5: Solve for \( d^2 \)
Rearranging gives:
\[ -10d^2 = 14 - 24 \implies -10d^2 = -10 \implies d^2 = 1 \]
Thus, \( d = 1 \) or \( d = -1 \).

### Step 6: Find the Roots
Substituting \( d = 1 \) into the roots:
- \( 2 - 3(1) = -1 \)
- \( 2 - 1 = 1 \)
- \( 2 + 1 = 3 \)
- \( 2 + 3(1) = 5 \)

Thus, the roots are:
\[ -1, 1, 3, 5 \]

### Final Answer
The roots of the equation \( x^4 - 8x^3 + 14x^2 + 8x - 15 = 0 \) are:
\[ \boxed{-1, 1, 3, 5} \]