Here are the solutions to the questions shown in the uploaded image:

Question 1: What are the zeros of the polynomial $6x^3 + 2x^2 - 4x$?

Solution:

1. Factor the polynomial: $6x^3 + 2x^2 - 4x = 2x(3x^2 + x - 2)$
2. Solve $3x^2 + x - 2 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \text{ where } a = 3, b = 1, c = -2$ $x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-2)}}{2(3)} = \frac{-1 \pm \sqrt{1 + 24}}{6} = \frac{-1 \pm \sqrt{25}}{6} = \frac{-1 \pm 5}{6}$ $x = \frac{-1 + 5}{6} = \frac{4}{6} = \frac{2}{3}, \quad x = \frac{-1 - 5}{6} = \frac{-6}{6} = -1$
3. The zeros are: $x = 0, x = \frac{2}{3}, x = -1$

Answer: (a) $0, \frac{2}{3}, -1$

Question 2: What does the graph below represent?

Solution:

1. The graph shows a parabola that opens downward.
2. This indicates a negative leading coefficient.
3. Since both ends of the graph go to $-\infty$, it must be an even-degree polynomial.

Answer: (a) Negative leading coefficient and even-degree polynomial

Question 3: Identify the degree, leading coefficient, and leading term of the polynomial $-2x^{10} - 25x^4 + 30x^3 - 20x + 5$.

Solution:

1. The degree of the polynomial is the highest power of $x$, which is $10$.
2. The leading term is the term with the highest degree: $-2x^{10}$.
3. The leading coefficient is the coefficient of the leading term, which is $-2$.

Answer: (d) $10, -2, -2x^{10}$

Question 4: Considering the removable discontinuities of the function $f(x) = \frac{x - 10}{x^2 - 12x + 20}$:

Solution:

1. Factor the denominator: $x^2 - 12x + 20 = (x - 10)(x - 2)$
2. The function becomes: $f(x) = \frac{x - 10}{(x - 10)(x - 2)}$
3. The removable discontinuity occurs where the numerator and denominator cancel, i.e., $x - 10 = 0 \Rightarrow x = 10$.
4. The non-removable discontinuity is at $x - 2 = 0 \Rightarrow x = 2$.

Answer: (c) $x = 10, 2$ are removable discontinuities

Question 5: What are the vertical asymptotes of the function $f(x) = \frac{x - 1}{x^2 + 4x + 3}$?

Solution:

1. Factor the denominator: $x^2 + 4x + 3 = (x + 1)(x + 3)$
2. The vertical asymptotes occur where the denominator is zero, i.e., $x + 1 = 0$ or $x + 3 = 0$. $x = -1, x = -3$

Answer: (a) $x = -3, x = -1$

Let me know if you'd like further details or have additional questions!

Expanding Questions:

1. How does factoring a polynomial help in finding its zeros?
2. What are the characteristics of even-degree vs. odd-degree polynomials in graphs?
3. How do you distinguish removable discontinuities from vertical asymptotes?
4. Why does a negative leading coefficient result in a downward-opening parabola?
5. How do you compute the vertical and horizontal asymptotes of a rational function?

Tip:

When solving polynomials or rational functions, always start by factoring completely—it simplifies finding roots, discontinuities, and asymptotes!