To solve the equation \( x^2(2x-1) + 3x(2x-1) - 4(2x-1) = 0 \), we can first factor out the common term, which is \( (2x-1) \):
\[ (2x-1)(x^2 + 3x - 4) = 0 \]
Now we can set each factor equal to zero:
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First, solve \( 2x - 1 = 0 \): \[ 2x = 1 \implies x = \frac{1}{2} \]
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Next, solve \( x^2 + 3x - 4 = 0 \). We can factor this quadratic: \[ x^2 + 3x - 4 = (x + 4)(x - 1) = 0 \] Setting each factor to zero gives: \[ x + 4 = 0 \implies x = -4 \] \[ x - 1 = 0 \implies x = 1 \]
Putting it all together, the complete set of solutions to the equation is: \[ x = \frac{1}{2}, -4, 1 \]
Therefore, the correct response is: \[ \frac{1}{2}, -4, 1 \]