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.<br /><br />7) $(x+1)(3-x) = 0$<br />$x+1 = 0$ or $3-x = 0$<br />$x = -1$ or $x = 3$<br /><br />10) $(x+7)(x-9) = 0$<br />$x+7 = 0$ or $x-9 = 0$<br />$x = -7$ or $x = 9$<br /><br />13) $(27-x)(x+9) = 0$<br />$27-x = 0$ or $x+9 = 0$<br />$x = 27$ or $x = -9$<br /><br />10) $(x-1)(x^2+1) = 0$<br />$x-1 = 0$ or $x^2+1 = 0$<br />$x = 1$ or $x^2 = -1$ (no real solution for x^2 = -1)<br />Therefore, only x = 1 is a solution.<br /><br />$(x^2+9)(x-2) = 0$<br />$x^2+9 = 0$ or $x-2 = 0$<br />$x^2 = -9$ (no real solution) or $x = 2$<br />Therefore, only x = 2 is a solution.<br /><br />8) $(8+x)(6-x) = 5$<br />This is not a simple zero-product property problem. We need to expand and solve a quadratic equation:<br />$48 - 8x + 6x - x^2 = 5$<br />$-x^2 - 2x + 48 - 5 = 0$<br />$-x^2 - 2x + 43 = 0$<br />$x^2 + 2x - 43 = 0$<br />Using the quadratic formula:<br />$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-43)}}{2(1)} = \frac{-2 \pm \sqrt{176}}{2} = -1 \pm \sqrt{44} = -1 \pm 2\sqrt{11}$<br /><br />11) $(5-x)(x+7) = 5$<br />This is also not a simple zero-product property problem. Expand and solve the quadratic:<br />$35 + 35x - x^2 - 7x = 5$<br />$-x^2 + 2x + 30 = 0$<br />$x^2 - 2x - 30 = 0$<br />Using the quadratic formula:<br />$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-30)}}{2} = \frac{2 \pm \sqrt{124}}{2} = 1 \pm \sqrt{31}$<br /><br />14) $(15-3x)(x-4) = 0$<br />$15-3x = 0$ or $x-4 = 0$<br />$3x = 15$ or $x = 4$<br />$x = 5$ or $x = 4$<br /><br />17) $(x^2+1)(5-x) = 0$<br />$x^2+1 = 0$ or $5-x = 0$<br />$x^2 = -1$ (no real solution) or $x = 5$<br />Therefore, only x = 5 is a solution.<br /><br />20) $(x^2+5)(x-5) = 0$<br />$x^2+5 = 0$ or $x-5 = 0$<br />$x^2 = -5$ (no real solution) or $x = 5$<br />Therefore, only x = 5 is a solution.<br /><br />gi $(x-3)(x+2) - 40 = 0$<br />$x^2 - x - 6 - 40 = 0$<br />$x^2 - x - 46 = 0$<br />Using the quadratic formula:<br />$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-46)}}{2} = \frac{1 \pm \sqrt{185}}{2}$<br /><br />(2) $(x-4)(x+5) = 0$<br />$x-4 = 0$ or $x+5 = 0$<br />$x = 4$ or $x = -5$<br /><br />$(6x+12)(-x-3) = 0$<br />$6(x+2)(-1)(x+3) = 0$<br />$x = -2$ or $x = -3$<br /><br />$(x+3)(x^2+1) = 0$<br />$x+3 = 0$ or $x^2+1 = 0$<br />$x = -3$ or $x^2 = -1$ (no real solution)<br />Therefore, only x = -3 is a solution.<br /><br />$(2+x)(x^2+2) = 0$<br />$2+x = 0$ or $x^2+2 = 0$<br />$x = -2$ or $x^2 = -2$ (no real solution)<br />Therefore, only x = -2 is a solution.<br /><br /><br />