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Factor: 4x2 + 8x + 3 (2x + 1)(2x + 3) (4x - 1)(x + 2) (4x + 3)(x + 1) prime
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Can someone help me Solve the equation by completing the square. 4x2 − 8x + 1 = 0
\[4(x^2-2x+\frac{1}{4}) \]\[4[(x-\frac{2}{2})^2-(\frac{2}{2})^2+\frac{1}{4}]\]\[4[ (x-1)^2-1+\frac{1}{4}]\]\[4[(x-1)^2-\frac{3}{4}] = 4(x-1)^2 -3\]
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