Find all possible values of \( b: x^{2}+b x-36 \)

We want to find all values of \( b \) such that the quadratic \[ x^2 + bx - 36 \] can be factored into two linear factors with integer coefficients. If the quadratic factors as \[ (x+m)(x+n), \] then by expanding we have: \[ x^2 + (m+n)x + mn. \] Comparing coefficients with \[ x^2 + bx - 36, \] we get the system: \[ m+n=b \quad \text{and} \quad mn=-36. \] To find the possible values of \( b \), we will first list all pairs of integers \( (m,n) \) satisfying \[ mn=-36. \] The integer pairs (considering order) that satisfy \( mn=-36 \) are: - \( m = 1 \) and \( n = -36 \) giving \( m+n = 1 + (-36) = -35 \). - \( m = -1 \) and \( n = 36 \) giving \( m+n = -1 + 36 = 35 \). - \( m = 2 \) and \( n = -18 \) giving \( m+n = 2 + (-18) = -16 \). - \( m = -2 \) and \( n = 18 \) giving \( m+n = -2 + 18 = 16 \). - \( m = 3 \) and \( n = -12 \) giving \( m+n = 3 + (-12) = -9 \). - \( m = -3 \) and \( n = 12 \) giving \( m+n = -3 + 12 = 9 \). - \( m = 4 \) and \( n = -9 \) giving \( m+n = 4 + (-9) = -5 \). - \( m = -4 \) and \( n = 9 \) giving \( m+n = -4 + 9 = 5 \). - \( m = 6 \) and \( n = -6 \) giving \( m+n = 6 + (-6) = 0 \). Thus, the corresponding possible values for \( b = m+n \) are: \[ -35, \, 35, \, -16, \, 16, \, -9, \, 9, \, -5, \, 5, \, 0. \] So, all possible values of \( b \) are: \[ \{ -35, 35, -16, 16, -9, 9, -5, 5, 0 \}. \]