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

Answer 1

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

\[ (2x-1)(x^2 + 3x - 4) = 0 \]

This gives us two equations to solve:

1. \( 2x - 1 = 0 \)
2. \( x^2 + 3x - 4 = 0 \)

Starting with the first equation:

\[ 2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \]

Next, we solve the second equation \( x^2 + 3x - 4 = 0 \). To factor this quadratic equation, we look for two numbers that multiply to \(-4\) and add to \(3\). The numbers \(4\) and \(-1\) work because:

\[ 4 + (-1) = 3 \quad \text{and} \quad 4 \cdot (-1) = -4 \]

Thus, we can factor the equation as follows:

\[ (x + 4)(x - 1) = 0 \]

Setting each factor equal to zero gives:

\[ x + 4 = 0 \Rightarrow x = -4 \]

\[ x - 1 = 0 \Rightarrow x = 1 \]

So the complete set of solutions from the two equations is:

\[ x = \frac{1}{2}, \quad x = -4, \quad x = 1 \]

Now we can compile the complete set of solutions:

\[ x = \frac{1}{2}, -4, 1 \]

Thus, the complete set of solutions to Janet's equation is \(\frac{1}{2}, -4, 1\). According to the response options provided, this corresponds to:

\(\frac{1}{2}, -4, 1\) (hidden in the format used).

So the answer is:

\(\frac{1}{2}, -4, 1\)