Fill in the missing factors to complete each factorization. \( \begin{array}{ll}x^{2}-3 x-4=(x+1)( & ) \\ \\ x^{2}-5 x+4=(x-4)(\quad) & x-1 \\ x^{2}+3 x-4=(x-1)( & ) \\ x^{2}+5 x+4=(x+4)(\quad) & x+1 \\ \end{array} \)

To complete the factorizations, we need to find the missing factors for each expression. Let's go through each one step by step. 1. **For \( x^{2}-3x-4=(x+1)(\quad) \)**: - We need to find a factor that, when multiplied by \( (x + 1) \), gives us \( x^2 - 3x - 4 \). - We can use the fact that \( (x + 1)(x + b) = x^2 + (1 + b)x + b \). - Setting \( 1 + b = -3 \) gives \( b = -4 \). - Therefore, the missing factor is \( (x - 4) \). 2. **For \( x^{2}-5x+4=(x-4)(\quad) \)**: - We need to find a factor that, when multiplied by \( (x - 4) \), gives us \( x^2 - 5x + 4 \). - We can use the fact that \( (x - 4)(x + b) = x^2 + (-4 + b)x - 4b \). - Setting \( -4 + b = -5 \) gives \( b = -1 \). - Therefore, the missing factor is \( (x - 1) \). 3. **For \( x^{2}+3x-4=(x-1)(\quad) \)**: - We need to find a factor that, when multiplied by \( (x - 1) \), gives us \( x^2 + 3x - 4 \). - We can use the fact that \( (x - 1)(x + b) = x^2 + (-1 + b)x - b \). - Setting \( -1 + b = 3 \) gives \( b = 4 \). - Therefore, the missing factor is \( (x + 4) \). 4. **For \( x^{2}+5x+4=(x+4)(\quad) \)**: - We need to find a factor that, when multiplied by \( (x + 4) \), gives us \( x^2 + 5x + 4 \). - We can use the fact that \( (x + 4)(x + b) = x^2 + (4 + b)x + 4b \). - Setting \( 4 + b = 5 \) gives \( b = 1 \). - Therefore, the missing factor is \( (x + 1) \). Now, we can fill in the missing factors: \[ \begin{array}{ll} x^{2}-3x-4=(x+1)(x-4) \\ \\ x^{2}-5x+4=(x-4)(x-1) \\ \\ x^{2}+3x-4=(x-1)(x+4) \\ \\ x^{2}+5x+4=(x+4)(x+1) \\ \end{array} \]