To factor the polynomial [tex]\( 12y^2 + 5y - 2 \)[/tex] completely, we need to find pairs of binomials that when multiplied give us the original polynomial. Let's go through the steps:
1. Identify the Polynomial:
We need to factor [tex]\( 12y^2 + 5y - 2 \)[/tex].
2. Choose the Factoring Method:
We will use trial and error with the provided options to check which set of binomials, when multiplied, gives us the original polynomial.
Let's start with the given options:
Option A: [tex]\((4y - 1)(3y + 2)\)[/tex]
Multiply [tex]\( (4y - 1) \)[/tex] and [tex]\( (3y + 2) \)[/tex]:
[tex]\[
(4y - 1) (3y + 2) = 4y \cdot 3y + 4y \cdot 2 - 1 \cdot 3y - 1 \cdot 2
\][/tex]
[tex]\[
= 12y^2 + 8y - 3y - 2
\][/tex]
[tex]\[
= 12y^2 + 5y - 2
\][/tex]
This matches the polynomial [tex]\( 12y^2 + 5y - 2 \)[/tex].
Clearly, option A is correct.
For completeness, let's briefly check the other options to ensure they do not give us the correct polynomial:
Option B: [tex]\((6y - 1)(2y + 2)\)[/tex]
[tex]\[
(6y - 1)(2y + 2) = 12y^2 + 12y - 2y - 2
\][/tex]
[tex]\[
= 12y^2 + 10y - 2
\][/tex]
This does not match the polynomial [tex]\( 12y^2 + 5y - 2 \)[/tex].
Option C: [tex]\((4y - 2)(3y + 1)\)[/tex]
[tex]\[
(4y - 2)(3y + 1) = 12y^2 + 4y - 6y - 2
\][/tex]
[tex]\[
= 12y^2 - 2y - 2
\][/tex]
This does not match the polynomial [tex]\( 12y^2 + 5y - 2 \)[/tex].
Option D: [tex]\((4y + 1)(3y - 2)\)[/tex]
[tex]\[
(4y + 1)(3y - 2) = 12y^2 - 8y + 3y - 2
\][/tex]
[tex]\[
= 12y^2 - 5y - 2
\][/tex]
This does not match the polynomial [tex]\( 12y^2 + 5y - 2 \)[/tex].
After evaluating all the given options, the correct factorization of [tex]\( 12y^2 + 5y - 2 \)[/tex] is indeed:
[tex]\[ \boxed{(4y - 1)(3y + 2)} \][/tex]
