The factors of `x^4 - 14x^2y^(2) - 51y^4`

Options

A. `(x+sqrt17y)(x-sqrt17y)(x^2+3y^2)`
B. `(x+sqrt7y)(x-sqrt17y)(x^2+3y^2)`
C. `(x+sqrt17y)(x-sqrt17y)(x^2+y^2)`
D. None of the above

Answer

To factor the polynomial \( x^4 - 14x^2y^2 - 51y^4 \), we can follow these steps:

### Step 1: Rewrite the polynomial
We start with the polynomial:
\[ x^4 - 14x^2y^2 - 51y^4 \]
Notice that we can treat \( x^2 \) as a single variable. Let \( z = x^2 \). Then the polynomial becomes:
\[ z^2 - 14zy^2 - 51y^4 \]

### Step 2: Identify the coefficients
In the quadratic form \( z^2 + bz + c \), we have:
- \( b = -14y^2 \)
- \( c = -51y^4 \)

### Step 3: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( ac = 1 \cdot (-51y^4) = -51y^4 \) and add to \( b = -14y^2 \).

The two numbers that satisfy this are \( -17y^2 \) and \( 3y^2 \) because:
\[ -17y^2 + 3y^2 = -14y^2 \]
and
\[ -17y^2 \cdot 3y^2 = -51y^4 \]

### Step 4: Rewrite the middle term
Now we can rewrite the polynomial:
\[ z^2 - 17zy^2 + 3zy^2 - 51y^4 \]

### Step 5: Factor by grouping
Group the terms:
\[ (z^2 - 17zy^2) + (3zy^2 - 51y^4) \]
Factor out the common factors:
\[ z(z - 17y^2) + 3y^2(z - 17y^2) \]
Now factor out \( (z - 17y^2) \):
\[ (z - 17y^2)(z + 3y^2) \]

### Step 6: Substitute back for \( z \)
Now substitute back \( z = x^2 \):
\[ (x^2 - 17y^2)(x^2 + 3y^2) \]

### Step 7: Final factorization
Thus, the final factorization of the polynomial \( x^4 - 14x^2y^2 - 51y^4 \) is:
\[ (x^2 - 17y^2)(x^2 + 3y^2) \]