The product of two numbers is `16x^(4) -1`. If one number is 2x - 1, find the other.

Answer

To find the other number when the product of two numbers is given as \( 16x^4 - 1 \) and one number is \( 2x - 1 \), we can follow these steps:

### Step 1: Set up the equation
Let \( a = 2x - 1 \) and \( b \) be the other number. We know that:
\[ a \cdot b = 16x^4 - 1 \]
Substituting \( a \) into the equation gives:
\[ (2x - 1) \cdot b = 16x^4 - 1 \]

### Step 2: Express \( 16x^4 - 1 \) in a different form
Notice that \( 16x^4 - 1 \) can be factored using the difference of squares:
\[ 16x^4 - 1 = (4x^2)^2 - 1^2 = (4x^2 - 1)(4x^2 + 1) \]

### Step 3: Factor \( 4x^2 - 1 \)
The term \( 4x^2 - 1 \) can also be factored further using the difference of squares:
\[ 4x^2 - 1 = (2x - 1)(2x + 1) \]
Thus, we can rewrite \( 16x^4 - 1 \) as:
\[ (2x - 1)(2x + 1)(4x^2 + 1) \]

### Step 4: Substitute back into the equation
Now we have:
\[ (2x - 1) \cdot b = (2x - 1)(2x + 1)(4x^2 + 1) \]

### Step 5: Solve for \( b \)
Since \( 2x - 1 \) is common on both sides, we can divide both sides by \( 2x - 1 \) (assuming \( 2x - 1 \neq 0 \)):
\[ b = (2x + 1)(4x^2 + 1) \]

### Step 6: Expand \( b \)
Now we need to expand \( b \):
\[ b = (2x + 1)(4x^2 + 1) = 2x \cdot 4x^2 + 2x \cdot 1 + 1 \cdot 4x^2 + 1 \cdot 1 \]
\[ = 8x^3 + 2x + 4x^2 + 1 \]

### Step 7: Rearranging the terms
Rearranging the terms gives us:
\[ b = 8x^3 + 4x^2 + 2x + 1 \]

Thus, the other number \( b \) is:
\[ \boxed{8x^3 + 4x^2 + 2x + 1} \]