Choose the correct answer in each drop-down list to complete the factorization: \( x^{2}-4 x-12=(x- \) Select \( -\vee)(x- \) Select \( -\vee) \) \( 4 x^{2}+11 x-20=(- \) Select \( -\vee-5)(- \) Select \( -\vee+4) \) \( 3 x^{2}+2 x-16=(3 x- \) Select \( -\vee)(x- \) Select \( -\vee) \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-4x-12=0\) - step1: Factor the expression: \(\left(x-6\right)\left(x+2\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-6=0\\&x+2=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=6\\&x=-2\end{align}\) - step4: Rewrite: \(x_{1}=-2,x_{2}=6\) Solve the equation \( 3 x^{2}+2 x-16=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(3x^{2}+2x-16=0\) - step1: Factor the expression: \(\left(x-2\right)\left(3x+8\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-2=0\\&3x+8=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=2\\&x=-\frac{8}{3}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{8}{3},x_{2}=2\) Solve the equation \( 4 x^{2}+11 x-20=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(4x^{2}+11x-20=0\) - step1: Factor the expression: \(\left(x+4\right)\left(4x-5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&4x-5=0\\&x+4=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{5}{4}\\&x=-4\end{align}\) - step4: Rewrite: \(x_{1}=-4,x_{2}=\frac{5}{4}\) Let's solve each factorization step by step. 1. **For the equation \( x^{2}-4x-12=0 \)**: - The roots are \( x_{1} = -2 \) and \( x_{2} = 6 \). - Therefore, the factorization is: \[ x^{2}-4x-12 = (x - (-2))(x - 6) = (x + 2)(x - 6) \] 2. **For the equation \( 4x^{2}+11x-20=0 \)**: - The roots are \( x_{1} = -4 \) and \( x_{2} = \frac{5}{4} \) (or \( 1.25 \)). - Therefore, the factorization is: \[ 4x^{2}+11x-20 = (4x - (-4))(4x - \frac{5}{4}) = (4x + 4)(4x - 5) \] 3. **For the equation \( 3x^{2}+2x-16=0 \)**: - The roots are \( x_{1} = -\frac{8}{3} \) and \( x_{2} = 2 \). - Therefore, the factorization is: \[ 3x^{2}+2x-16 = (3x - (-\frac{8}{3}))(x - 2) = (3x + \frac{8}{3})(x - 2) \] Now, let's fill in the blanks: 1. \( x^{2}-4x-12=(x+2)(x-6) \) 2. \( 4x^{2}+11x-20=(4x+4)(4x-5) \) 3. \( 3x^{2}+2x-16=(3x+\frac{8}{3})(x-2) \) So the completed factorization is: - \( x^{2}-4x-12=(x+2)(x-6) \) - \( 4x^{2}+11x-20=(4x+4)(4x-5) \) - \( 3x^{2}+2x-16=(3x+\frac{8}{3})(x-2) \)