Solve for x: 1.1.1 (x−2)(3x+1)(x+5)=0 1.1.2 5x2−x−9=0 (correct answer to two decimal places) 1.1.3 sqrt(x−2)+x=4 1.1.4 x2+7x<=-6 1.1.5 x^(5/3)=32

For the product of factors to be zero, at least one of the factors must be zero.

$$(x-2)(3x+1)(x+5) = 0$$

This means:

$$x-2 = 0 \quad \text{or} \quad 3x+1 = 0 \quad \text{or} \quad x+5 = 0$$

From the first equation:

$$x-2 = 0$$ $$x = 2$$

From the second equation:

$$3x+1 = 0$$ $$3x = -1$$ $$x = -\frac{1}{3}$$

From the third equation:

$$x+5 = 0$$ $$x = -5$$

The solutions for $x$ are:
$x = 2 \text{ or } x = -\frac{1}{3} \text{ or } x = -5$