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Factorizacion \( 1=12 x^{3}-18 x^{2}+6 x \)

Ask by Huang Elliott. in Mexico
Dec 02,2024

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La factorización de \( 12x^3 - 18x^2 + 6x - 1 \) es \( 6x(2x - 1)(x - 1) = 0 \).

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¡Claro! Para factorizar el polinomio \( 12x^{3} - 18x^{2} + 6x \), primero puedes extraer el factor común. En este caso, el común es \( 6x \). Así que, sacando el factor, queda: \( 6x(2x^{2} - 3x + 1) \). ¡Espera! Ahora necesitas factorizar el trinomio \( 2x^{2} - 3x + 1 \). Esto se puede hacer encontrando los números que multiplicados dan \( 2 \) (el coeficiente de \( x^{2} \) y \( 1 \) (el término constante) y que sumados dan \( -3 \) (el coeficiente de \( x \)). Después de hacer ese análisis, encontramos que: \( 2x^{2} - 3x + 1 = (2x - 1)(x - 1) \). Por lo tanto, la factorización completa de \( 12x^{3} - 18x^{2} + 6x \) es: \( 6x(2x - 1)(x - 1) \). ¡Y listo! La factorización está completa.

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