Solve the equation by factoring. \( x^{2}+2 x-8=0 \)

To solve the equation \( x^{2}+2x-8=0 \) by factoring, we want to express this quadratic in the form of \( (x + p)(x + q) = 0 \). We need two numbers that multiply to -8 (the constant term) and add up to 2 (the coefficient of \( x \)). The numbers 4 and -2 fit these criteria because \( 4 \times -2 = -8 \) and \( 4 + (-2) = 2 \). So we can factor the equation as: \[ (x + 4)(x - 2) = 0 \] Now, we can set each factor to zero: 1. \( x + 4 = 0 \) which gives \( x = -4 \) 2. \( x - 2 = 0 \) which gives \( x = 2 \) Thus, the solutions to the equation \( x^{2}+2x-8=0 \) are \( x = -4 \) and \( x = 2 \).