Factor: Se l by group: \( x^{2}+4 x+3 x+20 \) \( 2 x^{2}+8 x^{2}-2 x \)

Sure, let's factor each of the given polynomial expressions by grouping. ### 1. \( x^{2} + 4x + 3x + 20 \) **Step 1: Combine like terms.** \[ x^{2} + 4x + 3x + 20 = x^{2} + 7x + 20 \] **Step 2: Attempt to factor by grouping.** We look for two numbers that multiply to \( 20 \) (the constant term) and add up to \( 7 \) (the coefficient of \( x \)). However, the pairs of factors of 20 are: - \( 1 \) and \( 20 \) (sum = \( 21 \)) - \( 2 \) and \( 10 \) (sum = \( 12 \)) - \( 4 \) and \( 5 \) (sum = \( 9 \)) None of these pairs add up to \( 7 \). **Conclusion:** The polynomial \( x^{2} + 7x + 20 \) **cannot be factored** using integer coefficients. --- ### 2. \( 2x^{2} + 8x^{2} - 2x \) **Step 1: Combine like terms.** \[ 2x^{2} + 8x^{2} - 2x = 10x^{2} - 2x \] **Step 2: Factor out the greatest common factor (GCF).** The GCF of \( 10x^{2} \) and \( -2x \) is \( 2x \). \[ 10x^{2} - 2x = 2x(5x - 1) \] **Conclusion:** The polynomial factors as \( 2x(5x - 1) \). --- **Final Summary:** 1. \( x^{2} + 4x + 3x + 20 = x^{2} + 7x + 20 \) **cannot be factored** over the integers. 2. \( 2x^{2} + 8x^{2} - 2x = 2x(5x - 1) \)