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5. Factorice completamente las siguientes expresiones. 1 a) \( x^{4}+x-x^{3} y-y \) b) \( x^{3}-x-x^{2} y+y \) c) \( 6 x^{2}+x y-y^{2} \) d) \( a^{2}-b^{3}+2 b^{3} x^{2}-2 a^{2} x^{2} \) e) \( a^{2}+9 a+20 \) f) \( a^{2}-7 a+12 \) g) \( a^{2}-6 a+9 \) h) \( 6 x^{2}-x-2 \) i) \( 6 x^{2}+7 x y-3 y^{2} \) j) \( m^{4}+m^{2} n^{2}+n^{4} \) k) \( 15+14 x-8 x^{2} \) l) \( x^{6}+x^{3}-2 \) m) \( 2 \sqrt[3]{x^{2}}+5 \sqrt[3]{x}+2 \) n) \( 4 a^{2 n}-b^{2} \) n) \( x^{8}-y^{8} \) o) \( (\sqrt{6}-\sqrt{2}) \)

Ask by Davies Lambert. in Colombia
Mar 15,2025

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a) \( x^{4}+x-x^{3} y-y = (x^{2}-x+1)(x-y)(x+1) \) b) \( x^{3}-x-x^{2} y+y = (x-y)(x+1)(x-1) \) c) \( 6 x^{2}+x y-y^{2} = (3x-y)(2x+y) \) d) \( a^{2}-b^{3}+2 b^{3} x^{2}-2 a^{2} x^{2} = (b^{3}-a^{2})(2x^{2}-1) \) e) \( a^{2}+9 a+20 = (a+4)(a+5) \) f) \( a^{2}-7 a+12 = (a-4)(a-3) \) g) \( a^{2}-6 a+9 = (a-3)^{2} \) h) \( 6 x^{2}-x-2 = (2x+1)(3x-2) \) i) \( 6 x^{2}+7 x y-3 y^{2} = (3x-y)(2x+3y) \) j) \( m^{4}+m^{2} n^{2}+n^{4} = (m^{2}+mn+n^{2})(m^{2}-mn+n^{2}) \) k) \( 15+14 x-8 x^{2} = (5-2x)(3+4x) \) l) \( x^{6}+x^{3}-2 = (x-1)(x^{2}+x+1)(x^{3}+2) \) m) \( 2 \sqrt[3]{x^{2}}+5 \sqrt[3]{x}+2 = (\sqrt[3]{x}+2)(2\sqrt[3]{x}+1) \) n) \( 4 a^{2 n}-b^{2} = (2a^{n}+b)(2a^{n}-b) \) o) \( x^{8}-y^{8} = (x-y)(x+y)(x^{2}+y^{2})(x^{4}+y^{4}) \) p) \( (\sqrt{6}-\sqrt{2}) \) no se puede factorizar más.

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Para la expresión \( x^{4}+x-x^{3} y-y \), primero Agrupamos los términos: \( (x^{4} - x^{3}y) + (x - y) \). Sacamos factor común en la primera parte: \( x^{3}(x - y) + 1(x - y) \), y obtenemos \( (x - y)(x^3 + 1) \). Luego, factorizamos \( x^3 + 1 \) como \( (x + 1)(x^2 - x + 1) \). Entonces, la expresión completa es \( (x - y)(x + 1)(x^2 - x + 1) \). Para la expresión \( 6 x^{2}+7 x y-3 y^{2} \), notamos que tiene la forma de un trinomio cuadrático. Usamos la fórmula general o factorización por prueba y error, que nos puede llevar a los factores \( (3x - y)(2x + 3y) \). Verificando la multiplicación de estos factores, confirmamos que son correctos.

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