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Tema III: Factoriza la suma y la diferencia de los siguientes cuadrados.

a) [tex]\(x^2 + 6x + 9\)[/tex]

b) [tex]\(x^2 - 30x + 225\)[/tex]

c) [tex]\(4x^2 + 12x + 9\)[/tex]

d) [tex]\(9x^2 - 12xy + 4y^2\)[/tex]

Sagot :

Vamos a factorizar cada una de las expresiones dadas, paso a paso.

### Parte (a):
[tex]\[ x^2 + 6x + 9 \][/tex]

Este es un trinomio cuadrado perfecto, ya que se ajusta a la forma de [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex].

Notemos que aquí:
- [tex]\(a = x\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(2ab = 2(x)(3) = 6x\)[/tex]
- [tex]\(b^2 = 3^2 = 9\)[/tex]

Entonces,
[tex]\[ x^2 + 6x + 9 = (x + 3)^2 \][/tex]

### Parte (b):
[tex]\[ x^2 - 30x + 225 \][/tex]

De nuevo, tenemos un trinomio cuadrado perfecto, de la forma [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex].

Notemos que:
- [tex]\(a = x\)[/tex]
- [tex]\(b = 15\)[/tex]
- [tex]\(2ab = 2(x)(15) = 30x\)[/tex]
- [tex]\(b^2 = 15^2 = 225\)[/tex]

Por lo tanto,
[tex]\[ x^2 - 30x + 225 = (x - 15)^2 \][/tex]

### Parte (c):
[tex]\[ 4x^2 + 12x + 9 \][/tex]

Este también es un trinomio cuadrado perfecto de la forma [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex].

Aquí:
- [tex]\(a = 2x\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(2ab = 2(2x)(3) = 12x\)[/tex]
- [tex]\(b^2 = 3^2 = 9\)[/tex]

Entonces,
[tex]\[ 4x^2 + 12x + 9 = (2x + 3)^2 \][/tex]

### Parte (d):
[tex]\[ 9x^2 - 12xy + 4y^2 \][/tex]

Esta expresión es un trinomio cuadrado perfecto de la forma [tex]\((a - b)^2 = a^2 - 2ab + b^2\)[/tex].

En este caso:
- [tex]\(a = 3x\)[/tex]
- [tex]\(b = 2y\)[/tex]
- [tex]\(2ab = 2(3x)(2y) = 12xy\)[/tex]
- [tex]\(b^2 = (2y)^2 = 4y^2\)[/tex]

Por lo tanto,
[tex]\[ 9x^2 - 12xy + 4y^2 = (3x - 2y)^2 \][/tex]

### Resumen de las factorizaciones:
a) [tex]\[ x^2 + 6x + 9 = (x + 3)^2 \][/tex]
b) [tex]\[ x^2 - 30x + 225 = (x - 15)^2 \][/tex]
c) [tex]\[ 4x^2 + 12x + 9 = (2x + 3)^2 \][/tex]
d) [tex]\[ 9x^2 - 12xy + 4y^2 = (3x - 2y)^2 \][/tex]
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