Let's solve the given problems step by step:

Given:

$f(x) = 4x^3 - 6x^2 + 2x$

i. Express $f(x)$ as a product of its first derivatives.

To find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(4x^3 - 6x^2 + 2x) = 12x^2 - 12x + 2$

Now, let's try to factorize $f(x)$ using common factors: $f(x) = 2x(2x^2 - 3x + 1)$

The quadratic can be further factorized: $2x(2x^2 - 3x + 1) = 2x(2x - 1)(x - 1)$

So, the factorized form of $f(x)$ is: $f(x) = 2x(2x - 1)(x - 1)$

ii. Solve the equation $f(x) = 0$.

Using the factorized form: $2x(2x - 1)(x - 1) = 0$

Setting each factor equal to zero gives:

• $2x = 0 \Rightarrow x = 0$
• $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
• $x - 1 = 0 \Rightarrow x = 1$

So, the solutions are: $x = 0, \quad x = \frac{1}{2}, \quad x = 1$

iii. Given $h(x) = 2x^2 - 3x - 9$. Find the values of $x$ for which $h(x) < 1$.

We need to solve: $2x^2 - 3x - 9 < 1$ $2x^2 - 3x - 10 < 0$

Let's find the roots of the quadratic equation $2x^2 - 3x - 10 = 0$ using the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2}$ $x = \frac{3 \pm \sqrt{9 + 80}}{4}$ $x = \frac{3 \pm \sqrt{89}}{4}$

The approximate solutions are: $x_1 = \frac{3 - \sqrt{89}}{4} \quad \text{and} \quad x_2 = \frac{3 + \sqrt{89}}{4}$

Since this is a parabola opening upwards (coefficient of $x^2$ is positive), the inequality $2x^2 - 3x - 10 < 0$ holds between the roots: $\frac{3 - \sqrt{89}}{4} < x < \frac{3 + \sqrt{89}}{4}$

Would you like a more detailed explanation of any part? Here are some related questions to explore:

1. How do you factor polynomials?
2. What is the quadratic formula and how is it used?
3. How do you determine intervals for which a quadratic inequality is satisfied?
4. What is the relationship between the derivative and the factorization of a polynomial?
5. How can you graphically represent the solution to an inequality?

Tip: When solving inequalities involving quadratic expressions, remember to find the roots and test intervals between them.