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 \( (2x - 1) \):
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Factor out \( (2x - 1) \): \[ (2x - 1)(x^2 + 3x - 4) = 0 \]
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Now we can solve for \( 2x - 1 = 0 \): \[ 2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2} \]
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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: \[ x^2 + 3x - 4 = (x + 4)(x - 1) \]
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Setting each factor to zero gives:
- \( x + 4 = 0 \implies x = -4 \)
- \( x - 1 = 0 \implies x = 1 \)
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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: \[ \text{12, −4, 1} \]
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