The equation whose roots exceed by 2 than the roots of ` 4x^4 + 32x^3 + 83 x^2 + 76 x + 21 =0` is

Options

A. `4x^4 + 13 x^2 + 9 =0`
B. `4x^4 -13x^2 +9=0`
C. ` 4x^4 +12 x^2 -9=0`
D. `4x^4 + 12x^2 -9=0`

Answer

To find the equation whose roots exceed those of the polynomial \( 4x^4 + 32x^3 + 83x^2 + 76x + 21 = 0 \) by 2, we can follow these steps:

### Step 1: Identify the given polynomial
The given polynomial is:
\[ 4x^4 + 32x^3 + 83x^2 + 76x + 21 = 0 \]

### Step 2: Substitute the roots
Let the roots of the given polynomial be \( r_1, r_2, r_3, r_4 \). According to the problem, we need to find a new polynomial whose roots are \( r_1 + 2, r_2 + 2, r_3 + 2, r_4 + 2 \).

### Step 3: Change of variable
To find the new polynomial, we can perform a change of variable. Let:
\[ y = x - 2 \quad \Rightarrow \quad x = y + 2 \]
Now, we substitute \( x = y + 2 \) into the original polynomial.

### Step 4: Substitute into the polynomial
Substituting \( x = y + 2 \) into the polynomial:
\[ 4(y + 2)^4 + 32(y + 2)^3 + 83(y + 2)^2 + 76(y + 2) + 21 = 0 \]

### Step 5: Expand the polynomial
Now we need to expand each term:
1. \( (y + 2)^4 = y^4 + 8y^3 + 24y^2 + 32y + 16 \)
2. \( (y + 2)^3 = y^3 + 6y^2 + 12y + 8 \)
3. \( (y + 2)^2 = y^2 + 4y + 4 \)
4. \( (y + 2) = y + 2 \)

Now substituting these expansions back into the polynomial:
\[ 4(y^4 + 8y^3 + 24y^2 + 32y + 16) + 32(y^3 + 6y^2 + 12y + 8) + 83(y^2 + 4y + 4) + 76(y + 2) + 21 = 0 \]

### Step 6: Combine like terms
Now combine all the terms:
- For \( y^4 \): \( 4y^4 \)
- For \( y^3 \): \( 32y^3 + 32y^3 = 64y^3 \)
- For \( y^2 \): \( 96y^2 + 83y^2 = 179y^2 \)
- For \( y \): \( 128y + 76y = 204y \)
- Constant term: \( 64 + 21 = 85 \)

Putting it all together, we get:
\[ 4y^4 + 64y^3 + 179y^2 + 204y + 85 = 0 \]

### Step 7: Write the final polynomial
Thus, the polynomial whose roots exceed those of the original polynomial by 2 is:
\[ 4y^4 + 64y^3 + 179y^2 + 204y + 85 = 0 \]