1) xcdot (2-x)=0 2) 8xcdot (-5-x)=d (x-2)cdot (x+1)=0 5) (x-1)cdot (x+2)=0 (-x)cdot (x-43)=0 6 (3-x)(x+4)=0 7) (x+1)cdot (3-x)=0 8) (8+x)cdot (6-x)=0 9) (x-3)(x+2)-(0)=0 (x+7)cdot (x-9)=0 (5-x)cdot (x+7)=0 12 (x-4)cdot (x+5)=0 3) (27-x)cdot (x+9)=0 (15-3.x)cdot (x-4)=0 (6x+12)cdot (-x-3)=0 (x-1)cdot (x^2+1)=0 (x^2+1)cdot (5-x)=0 (x+3)cdot (x^2+1)=0 (x^2+9)cdot (x-2)=0 20) (x^2+5)cdot (x-5)=0 (2+x)cdot (x^2+2)=0

The provided text is a list of equations to solve. Following the example, we'll find the values of x that make each equation true. Remember that if the product of two factors is zero, at least one of the factors must be zero.<br /><br />**Solutions:**<br /><br />1) x(2-x) = 0 => x = 0 or 2 - x = 0 => x = 0, x = 2<br />2) 8x(-5-x) = 0 => 8x = 0 or -5 - x = 0 => x = 0, x = -5<br />3) (-x)(x-43) = 0 => -x = 0 or x - 43 = 0 => x = 0, x = 43<br />4) (x-2)(x+1) = 0 => x - 2 = 0 or x + 1 = 0 => x = 2, x = -1<br />5) (x-1)(x+2) = 0 => x - 1 = 0 or x + 2 = 0 => x = 1, x = -2<br />6) (3-x)(x+4) = 0 => 3 - x = 0 or x + 4 = 0 => x = 3, x = -4<br />7) (x+1)(3-x) = 0 => x + 1 = 0 or 3 - x = 0 => x = -1, x = 3<br />8) (8+x)(6-x) = 0 => 8 + x = 0 or 6 - x = 0 => x = -8, x = 6<br />9) (x-3)(x+2) = 0 => x - 3 = 0 or x + 2 = 0 => x = 3, x = -2<br />10) (x+7)(x-9) = 0 => x + 7 = 0 or x - 9 = 0 => x = -7, x = 9<br />11) (5-x)(x+7) = 0 => 5 - x = 0 or x + 7 = 0 => x = 5, x = -7<br />12) (x-4)(x+5) = 0 => x - 4 = 0 or x + 5 = 0 => x = 4, x = -5<br />13) (27-x)(x+9) = 0 => 27 - x = 0 or x + 9 = 0 => x = 27, x = -9<br />14) (15-3x)(x-4) = 0 => 15 - 3x = 0 or x - 4 = 0 => x = 5, x = 4<br />15) (6x+12)(-x-3) = 0 => 6x + 12 = 0 or -x - 3 = 0 => x = -2, x = -3<br />16) (x-1)(x²+1) = 0 => x - 1 = 0 or x² + 1 = 0 (x² = -1, no real solution) => x = 1<br />17) (x²+1)(5-x) = 0 => x² + 1 = 0 (no real solution) or 5 - x = 0 => x = 5<br />18) (x+3)(x²+1) = 0 => x + 3 = 0 or x² + 1 = 0 (no real solution) => x = -3<br />19) (x²+9)(x-2) = 0 => x² + 9 = 0 (no real solution) or x - 2 = 0 => x = 2<br />20) (x²+5)(x-5) = 0 => x² + 5 = 0 (no real solution) or x - 5 = 0 => x = 5<br />21) (2+x)(x²+2) = 0 => 2 + x = 0 or x² + 2 = 0 (no real solution) => x = -2<br /><br /><br />**Note:** Some equations have no real solutions because a sum of squares cannot equal zero unless each term is zero (and in these cases, one term is a positive constant). Only real solutions are presented above.<br />