To factor the quadratic expression [tex]\( 12 y^2 + 5 y - 2 \)[/tex] completely, we can follow these steps:
1. Identify the quadratic form:
The expression is in the standard quadratic form [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a = 12 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -2 \)[/tex].
2. Set up the factoring:
Since the quadratic expression [tex]\( 12 y^2 + 5 y - 2 \)[/tex] is factorable, it can be written as a product of two binomials:
[tex]\[
(m y + n)(p y + q)
\][/tex]
Here, the coefficients [tex]\( m \)[/tex], [tex]\( n \)[/tex], [tex]\( p \)[/tex], and [tex]\( q \)[/tex] need to be determined.
3. Find appropriate values:
We need to find two numbers that multiply to [tex]\( a \times c \)[/tex] (which is [tex]\( 12 \times -2 = -24 \)[/tex]) and add up to [tex]\( b \)[/tex] (which is [tex]\( 5 \)[/tex]). After checking possible factors, we find:
[tex]\[
\begin{aligned}
3y \times 4y &= 12 y^2 \\
3y \times (-1) + 4y \times 2 &= 3y - 4y = 5y \\
2 \times -1 &= -2 \\
\end{aligned}
\][/tex]
So, the numbers that work are [tex]\( 3y \)[/tex] and [tex]\( 4y \)[/tex], giving the factors:
[tex]\[
(3y + 2)(4y - 1)
\][/tex]
4. Confirm the factorization:
To ensure the factorization is correct, expand [tex]\( (3y + 2)(4y - 1) \)[/tex] to check:
[tex]\[
\begin{aligned}
(3y + 2)(4y - 1) &= 3y \cdot 4y + 3y \cdot (-1) + 2 \cdot 4y + 2 \cdot (-1) \\
&= 12 y^2 - 3 y + 8 y - 2 \\
&= 12 y^2 + 5 y - 2
\end{aligned}
\][/tex]
The expanded form matches the original quadratic expression.
Thus, the correct factorization of [tex]\( 12 y^2 + 5 y - 2 \)[/tex] is [tex]\((3 y + 2)(4 y - 1)\)[/tex], and the best answer from the provided choices is:
[tex]\[
\boxed{(4 y - 1)(3 y + 2)}
\][/tex] which corresponds to answer choice A.
