Question

Factor the polynomial completely
48x^2y-3y

Answer 1

To factor the polynomial \( 48x^2y - 3y \) completely, we can start by looking for a common factor in both terms.

1. **Identify the common factor:**
Both terms \( 48x^2y \) and \( -3y \) have \( y \) as a common factor.

2. **Factor out the common factor:**
\[
48x^2y - 3y = y(48x^2 - 3)
\]

3. **Factor the expression inside the parentheses if possible:**
Let's look at \( 48x^2 - 3 \).
We can factor out a 3 from this expression:
\[
48x^2 - 3 = 3(16x^2 - 1)
\]
Here, \( 16x^2 - 1 \) is a difference of squares, which can be factored further:
\[
16x^2 - 1 = (4x - 1)(4x + 1)
\]
So, we have:
\[
48x^2 - 3 = 3(4x - 1)(4x + 1)
\]

4. **Put it all together:**
Now substitute it back into the factorization we obtained previously:
\[
y(48x^2 - 3) = y \cdot 3(4x - 1)(4x + 1)
\]

Thus, the completely factored form of the polynomial \( 48x^2y - 3y \) is:
\[
\boxed{3y(4x - 1)(4x + 1)}
\]