Questions: Solve the quadratic equation by factoring. (Enter your answers as a comma-separated list.) 6x^2+2x=0 x= Solve the quadratic equation by factoring. (Enter your answers as comma-separated list.) 20x^2-5x=0 x= Solve the quadratic equation by factoring. (Enter your answers as a comma-separated list.) x^2+4x+4=0 x=

1. For the first equation \(6x^2 + 2x = 0\), factor out the greatest common factor, which is \(2x\), and then solve for \(x\) by setting each factor equal to zero.

2. For the second equation \(20x^2 - 5x = 0\), factor out the greatest common factor, which is \(5x\), and then solve for \(x\) by setting each factor equal to zero.

3. For the third equation \(x^2 + 4x + 4 = 0\), recognize it as a perfect square trinomial and factor it as \((x + 2)^2 = 0\), then solve for \(x\).

Para la ecuación \(6x^2 + 2x = 0\), factorizamos y encontramos las soluciones. Al factorizar, obtenemos: \[ 2x(3x + 1) = 0 \] De aquí, establecemos cada factor igual a cero:

1. \(2x = 0 \Rightarrow x = 0\)
2. \(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\)

Las soluciones son \(x = 0\) y \(x = -\frac{1}{3}\).

Para la ecuación \(20x^2 - 5x = 0\), también factorizamos: \[ 5x(4x - 1) = 0 \] Establecemos cada factor igual a cero:

1. \(5x = 0 \Rightarrow x = 0\)
2. \(4x - 1 = 0 \Rightarrow x = \frac{1}{4}\)

Las soluciones son \(x = 0\) y \(x = \frac{1}{4}\).

Para la ecuación \(x^2 + 4x + 4 = 0\), reconocemos que es un trinomio cuadrado perfecto: \[ (x + 2)^2 = 0 \] De aquí, tenemos: \[ x + 2 = 0 \Rightarrow x = -2 \]

La única solución es \(x = -2\).

Las soluciones son:

• Para la primera ecuación: \(x = 0, -\frac{1}{3}\)
• Para la segunda ecuación: \(x = 0, \frac{1}{4}\)
• Para la tercera ecuación: \(x = -2\)

\[ \boxed{x = 0, -\frac{1}{3}; \quad x = 0, \frac{1}{4}; \quad x = -2} \]