let `z_1,z_2,z_3` and `z_4` be the roots of the equation `z^4 + z^3 +2=0` , then the value of `prod_(r=1)^(4) (2z_r+1)` is equal to :

Options

A. 28
B. 29
C. 30
D. 31

Answer

To solve the problem, we need to find the value of the product \( \prod_{r=1}^{4} (2z_r + 1) \), where \( z_1, z_2, z_3, z_4 \) are the roots of the polynomial equation \( z^4 + z^3 + 2 = 0 \).

### Step 1: Substitute \( z \) with \( \frac{y - 1}{2} \)

To find the new roots \( 2z_r + 1 \), we can set \( y = 2z + 1 \). This implies that \( z = \frac{y - 1}{2} \).

### Step 2: Rewrite the original polynomial

Substituting \( z \) into the original polynomial:

\[ z^4 + z^3 + 2 = 0 \]

becomes:

\[ \left(\frac{y - 1}{2}\right)^4 + \left(\frac{y - 1}{2}\right)^3 + 2 = 0 \]

### Step 3: Expand the polynomial

Now, we need to expand this expression:

\[ \left(\frac{y - 1}{2}\right)^4 = \frac{(y - 1)^4}{16} \]
\[ \left(\frac{y - 1}{2}\right)^3 = \frac{(y - 1)^3}{8} \]

So, we have:

\[ \frac{(y - 1)^4}{16} + \frac{(y - 1)^3}{8} + 2 = 0 \]

### Step 4: Multiply through by 16 to eliminate the denominators

Multiplying the entire equation by 16 gives:

\[ (y - 1)^4 + 2(y - 1)^3 + 32 = 0 \]

### Step 5: Expand \( (y - 1)^4 \) and \( (y - 1)^3 \)

Now, we expand \( (y - 1)^4 \) and \( (y - 1)^3 \):

\[ (y - 1)^4 = y^4 - 4y^3 + 6y^2 - 4y + 1 \]
\[ 2(y - 1)^3 = 2(y^3 - 3y^2 + 3y - 1) = 2y^3 - 6y^2 + 6y - 2 \]

### Step 6: Combine the expansions

Combining these expansions:

\[ y^4 - 4y^3 + 6y^2 - 4y + 1 + 2y^3 - 6y^2 + 6y - 2 + 32 = 0 \]

This simplifies to:

\[ y^4 - 2y^3 + 0y^2 + 2y + 31 = 0 \]

### Step 7: Find the product of the roots

The product of the roots of the polynomial \( ay^4 + by^3 + cy^2 + dy + e = 0 \) is given by \( (-1)^n \frac{e}{a} \), where \( n \) is the degree of the polynomial. Here, \( a = 1 \) and \( e = 31 \), so:

\[ \text{Product of the roots} = (-1)^4 \frac{31}{1} = 31 \]

### Final Answer

Thus, the value of \( \prod_{r=1}^{4} (2z_r + 1) \) is \( 31 \).

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