Question

Janet is trying to solve an equation and gets stuck at x2(2x−1)+3x(2x−1)−4(2x−1)=0 Use factor by grouping to continue factoring. Which of the following is a complete set of solutions to the equation?(1 point) Responses 12, −4, 1 12, −4, 1 −4, 1 −4, 1 12, 4, −1 12, 4, −1 12 1 half

Answer 1

To solve the equation \( x^2(2x - 1) + 3x(2x - 1) - 4(2x - 1) = 0 \) using factor by grouping, we can start by factoring out the common term \( (2x - 1) \):

1. Factor out \( (2x - 1) \):
\[
(2x - 1)(x^2 + 3x - 4) = 0
\]

2. Now we can solve for \( 2x - 1 = 0 \):
\[
2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}
\]

3. Next, we factor the quadratic \( x^2 + 3x - 4 \):
- We look for two numbers that multiply to \(-4\) (the constant term) and add to \(3\) (the coefficient of \(x\)). These numbers are \(4\) and \(-1\).
- Thus, we can factor the quadratic:
\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\]

4. Setting each factor to zero gives:
- \( x + 4 = 0 \implies x = -4 \)
- \( x - 1 = 0 \implies x = 1 \)

5. Now we have the complete set of solutions:
\[
x = \frac{1}{2}, \quad x = -4, \quad x = 1
\]

Thus, the complete set of solutions to the equation is \( \frac{1}{2}, -4, 1 \).

Based on the given response options, the correct answer is:
\[
\text{12, −4, 1}
\]