9-12-2023 Class Chapter 4 Part 1

9-12-2023 Class: Chapter 4 Part 1

Welcome to the recap of our 9-12-2023 class, where we delved into the fascinating world of polynomial and rational functions! We covered a range of essential topics designed to build your skills in algebraic manipulation and problem-solving. Let's review what we learned:

Topics Covered:

• Domain of a Rational Function: Interval Notation

  Understanding the domain is crucial! Remember that the domain of a rational function excludes values that make the denominator zero. For example, given $f(x) = \frac{x-8}{x^2 + 16x + 64}$ and $g(x) = \frac{x+5}{x^2 - x - 30}$, we find the domain by setting the denominators not equal to zero.

  For $f(x)$, $x^2 + 16x + 64 \neq 0$ implies $(x+8)(x+8) \neq 0$, so $x \neq -8$. Thus, the domain is $(-\infty, -8) \cup (-8, \infty)$.

  For $g(x)$, $x^2 - x - 30 \neq 0$ implies $(x+5)(x-6) \neq 0$, so $x \neq -5$ and $x \neq 6$. Thus, the domain is $(-\infty, -5) \cup (-5, 6) \cup (6, \infty)$.

• Solving Quadratic Inequalities

  We examined how to solve quadratic inequalities, both in factored and unfactored forms. A key technique is to find the zeros of the quadratic and test intervals on a number line.

  For example, solve $x(x-7) > 0$. The zeros are $x = 0$ and $x = 7$. Testing intervals $(-\infty, 0)$, $(0, 7)$, and $(7, \infty)$ reveals the solution to be $(-\infty, 0) \cup (7, \infty)$.

  Another example: Solve $x^2 - 6x > -8$. First, rewrite it as $x^2 - 6x + 8 > 0$. Factoring gives $(x-4)(x-2) > 0$. Zeros are $x = 2$ and $x = 4$. The solution is $(-\infty, 2) \cup (4, \infty)$.

• Polynomial Long Division

  Mastering polynomial long division is essential for simplifying rational expressions and finding factors. We worked through several problem types.

  Example: $(8x^3 + 2x^2 + 7x + 2) \div (2x^2 + x)$. The quotient is $4x - 1$ and the remainder is $8x + 2$.

• Synthetic Division

  A shortcut for dividing by a linear factor $(x - c)$.

  Example: $(9x^3 + 6x^2 - 21x + 7) \div (3x - 1)$. After performing synthetic division, we can express this as $3x^2 + 3x - 6 + \frac{1}{3x-1}$.

• Remainder Theorem

  The Remainder Theorem states that if a polynomial $P(x)$ is divided by $x - c$, then the remainder is $P(c)$. This gives us a quick way to evaluate polynomials at specific values.

  Example: Let $P(x) = -x^4 + 2x^3 - 4x + 5$. To find $P(2)$, we can use synthetic division or direct substitution. Using the remainder theorem, $P(2) = -3$.

• Factor Theorem

  The Factor Theorem states that $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.

  Example: Is $(x-3)$ a factor of $x^4 - 2x^3 + x^2 - 8x - 12$? By synthetic division, the remainder is 0, so yes, it is a factor.

• Rational Zeros Theorem

  This theorem helps us find potential rational roots of a polynomial. The possible rational zeros are of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

  We then use synthetic division to test these potential zeros. Once a zero is found, we can factor the polynomial and repeat the process to find all zeros.

Keep practicing these concepts, and you'll master polynomial and rational functions in no time! Remember to review your notes and work through additional examples. You've got this!