Let's break down and solve each of the questions from the quiz:

1. Solve the following equations by factorization:

a. $x^2 - x = 12$

We rearrange the equation as: $x^2 - x - 12 = 0$ Factorizing this quadratic equation: $(x - 4)(x + 3) = 0$ Thus, the solutions are: $x = 4 \quad \text{or} \quad x = -3$

b. $x^2 + 30 = x$

Rearrange the equation: $x^2 - x + 30 = 0$ The discriminant $\Delta = (-1)^2 - 4(1)(30) = 1 - 120 = -119$, so there are no real solutions, only complex ones.

c. $2x^2 + 6x = -16$

Rearrange the equation: $2x^2 + 6x + 16 = 0$ Dividing by 2: $x^2 + 3x + 8 = 0$ The discriminant is $\Delta = 3^2 - 4(1)(8) = 9 - 32 = -23$, so there are no real solutions, only complex ones.

d. $4x^2 - 17 = x$

Rearrange the equation: $4x^2 - x - 17 = 0$ Using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-17)}}{2(4)}$ $x = \frac{1 \pm \sqrt{1 + 272}}{8} = \frac{1 \pm \sqrt{273}}{8}$ So, the solutions are: $x = \frac{1 + \sqrt{273}}{8} \quad \text{or} \quad x = \frac{1 - \sqrt{273}}{8}$

2. Find the x-intercept (abscisse à l’origine) of the following equations:

a. $2x^2 + 3x - 40 = 0$

Using the quadratic formula: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-40)}}{2(2)}$ $x = \frac{-3 \pm \sqrt{9 + 320}}{4} = \frac{-3 \pm \sqrt{329}}{4}$ The solutions are: $x = \frac{-3 + \sqrt{329}}{4} \quad \text{or} \quad x = \frac{-3 - \sqrt{329}}{4}$

b. $4x^2 + 8x - 12 = 0$

Dividing by 4: $x^2 + 2x - 3 = 0$ Factorizing: $(x - 1)(x + 3) = 0$ The solutions are: $x = 1 \quad \text{or} \quad x = -3$

c. $x^2 - 6x - 16 = 0$

Using the quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-16)}}{2(1)}$ [ x = \frac{6 \