7) (x+1)cdot (3-x)=0 10) (x+7)cdot (x-9)=0 13) (27-x)cdot (x+9)=0 10) (x-1)cdot (x^2+1)=0 (x^2+9)cdot (x-2)=0 8) (8+x)cdot (6-x)=5 ]1) (5-x)cdot (x+7)=5 14) (15-3cdot x)cdot (x-4)=0 17) (x^2+1)cdot (5-x)=0 20) (x^2+5)cdot (x-5)=0 gi (x-3)cdot (x+2)-40=0 (2) (x-4)cdot (x+5)=0 (6x+12)cdot (-x-3)=0 I (x+3)cdot (x^2+1)=0 (2+x)cdot (x^2+2)=0

Let's solve these equations. The strategy is to use the zero-product property: if A * B = 0, then A = 0 or B = 0.7) or or 10) or or 13) or or 10) or or (no real solution for x^2 = -1)Therefore, only x = 1 is a solution. or (no real solution) or Therefore, only x = 2 is a solution.8) This is not a simple zero-product property problem. We need to expand and solve a quadratic equation: Using the quadratic formula: 11) This is also not a simple zero-product property problem. Expand and solve the quadratic: Using the quadratic formula: 14) or or or 17) or (no real solution) or Therefore, only x = 5 is a solution.20) or (no real solution) or Therefore, only x = 5 is a solution.gi Using the quadratic formula: (2) or or or or or (no real solution)Therefore, only x = -3 is a solution. or or (no real solution)Therefore, only x = -2 is a solution.