To factor the quadratic expression \(12d^2 + 4d - 1\), we will look for two binomials of the form \((Ad + B)(Cd + D)\). When multiplied, these should reproduce the original quadratic expression.
First, we need to use trial and error to determine the coefficients \(A\), \(B\), \(C\), and \(D\) such that:
\[ (Ad + B)(Cd + D) = 12d^2 + 4d - 1 \]
Let's expand each of the given options to see which one matches \(12d^2 + 4d - 1\).
1. \( (6d + 1)(2d + 1) \)
\[ (6d + 1)(2d + 1) = 6d \cdot 2d + 6d \cdot 1 + 1 \cdot 2d + 1 \cdot 1 \]
\[ = 12d^2 + 6d + 2d + 1 \]
\[ = 12d^2 + 8d + 1 \]
2. \( (6d - 1)(2d + 1) \)
\[ (6d - 1)(2d + 1) = 6d \cdot 2d + 6d \cdot 1 - 1 \cdot 2d - 1 \cdot 1 \]
\[ = 12d^2 + 6d - 2d - 1 \]
\[ = 12d^2 + 4d - 1 \]
3. \( (6d - 1)(2d - 1) \)
\[ (6d - 1)(2d - 1) = 6d \cdot 2d + 6d \cdot (-1) - 1 \cdot 2d - 1 \cdot (-1) \]
\[ = 12d^2 - 6d - 2d + 1 \]
\[ = 12d^2 - 8d + 1 \]
4. \( (6d + 1)(2d - 1) \)
\[ (6d + 1)(2d - 1) = 6d \cdot 2d + 6d \cdot (-1) + 1 \cdot 2d - 1 \cdot 1 \]
\[ = 12d^2 - 6d + 2d - 1 \]
\[ = 12d^2 - 4d - 1 \]
From the expansions, we see that option 2 gives us the original expression:
\[ (6d - 1)(2d + 1) = 12d^2 + 4d - 1 \]
Therefore, the correct factorization is:
\[ (6d - 1)(2d + 1) \]
12.
Factor the expression.
12d2 + 4d – 1
(6d + 1)(2d + 1)
(6d – 1)(2d + 1)
(6d – 1)(2d – 1)
(6d + 1)(2d – 1)
1 answer