What are the factors of `32x^(4)-500x` ?

Options

A. `4x(2x-5)(4x^(2)+10x+25)`
B. `4x(2x-5)(4x^(2)+10x+5)`
C. `4x(2x-5)(x^(2)+10x+25)`
D. None the above

Answer

To factor the expression \(32x^4 - 500x\), we can follow these steps:

### Step 1: Factor out the common terms
The expression \(32x^4 - 500x\) has a common factor of \(4x\). We can factor this out:

\[ 32x^4 - 500x = 4x(8x^3 - 125) \]

### Step 2: Recognize the difference of cubes
Next, we notice that \(8x^3 - 125\) can be expressed as a difference of cubes. We can rewrite \(8x^3\) as \((2x)^3\) and \(125\) as \(5^3\). Thus, we have:

\[ 8x^3 - 125 = (2x)^3 - 5^3 \]

### Step 3: Apply the difference of cubes formula
The difference of cubes can be factored using the identity:

\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]

In our case, \(a = 2x\) and \(b = 5\). Applying the formula gives us:

\[ (2x - 5)((2x)^2 + (2x)(5) + 5^2) \]

### Step 4: Simplify the expression
Now we simplify the second factor:

\[ (2x)^2 = 4x^2, \quad (2x)(5) = 10x, \quad 5^2 = 25 \]

Putting it all together, we have:

\[ (2x - 5)(4x^2 + 10x + 25) \]

### Step 5: Combine everything
Now we can write the complete factorization of the original expression:

\[ 32x^4 - 500x = 4x(2x - 5)(4x^2 + 10x + 25) \]

### Final Answer
Thus, the factors of \(32x^4 - 500x\) are:

\[ 4x(2x - 5)(4x^2 + 10x + 25) \]

---