Polynomial Long Division, Factor Theorem, and Synthetic Division Study Guide

When performing polynomial long division, it is essential to include placeholders for every power of $x$ to ensure correct alignment.
Example Problem: Divide $x^6 + x^4 + x^2 + 1$ by $x^2 + 1$ .
- Rewrite the dividend with placeholders: $x^6 + 0x^5 + x^4 + 0x^3 + x^2 + 0x + 1$ .
- Divide the leading term: $x^6 \text{ divided by } x^2 = x^4$ .
- Multiply and subtract: $(x^6 + x^4) - (x^6 + x^4) = 0$ .
- Bring down remaining terms. Because the next terms are zeros, bring down the $x^2 + 0x + 1$ part.
- Divide the next leading term: $x^2 \text{ divided by } x^2 = 1$ .
- Multiply and subtract: $(x^2 + 1) - (x^2 + 1) = 0$ .
- Result: The quotient is $x^4 + 1$ . The remainder is $0$ .
If a remainder exists (for example, $10$ ), the final answer should be written in the form: $\text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}}$ . If the remainder is zero, there is no need to write that fraction.
Factoring and long division are different processes. If a question asks specifically for long division, factoring to find the result is incorrect methodology even if the answer is similar.

Definition: The Remainder Theorem states that if you divide a polynomial $P(x)$ by a binomial of the form $x - c$ , then the remainder is equal to $P(c)$ .
To find the remainder of $P(x)$ divided by $x + 2$ , you must evaluate $P(-2)$ .
Example calculation:
- For a specific function $P(x) = 3x^5 + 5x^4 - x^3 + 2x^2 + 7x + 3$ (implied coefficients from discussion), evaluating $P(-2)$ results in a remainder of $5$ .
- The quotient for such a division is noted as $3x^4 - x^3 - 2x^2 + 4x - 1$ .

Factoring: The process of writing a mathematical expression as a product of two or more expressions. These expressions are known as factors.
Multiplier/Factor: A product of expressions that divide the original polynomial exactly, resulting in a remainder of zero.
Solving: The procedure of finding the value(s) of a variable that makes an equation true.
Satisfying the Equation: A value of $x$ satisfies an equation if, when plugged into the equation, both sides of the equation are equal.
- Example: If $x - 3 = 1$ , the solution is $x = 4$ because $4 - 3 = 1$ is a true statement.
- Testing $x = 2$ is unsuccessful because $2 - 3 = -1$ , and $-1 \neq 1$ , meaning $x=2$ does not satisfy the equation.
Synonyms for Solutions: There are four terms used interchangeably for values that satisfy a polynomial equation when set to zero:
1. Solutions
2. Zeros
3. Roots
4. $x$ -intercepts

Definition of the Factor Theorem: The Factor Theorem connects the roots of a polynomial to its factors. It states that a polynomial $P(x)$ has a factor $(x - c)$ if and only if $P(c) = 0$ .
Relationship between Zero and Factor:
- If $c$ is a zero, then $(x - c)$ is a factor.
- If $(x - c)$ is a factor, then $P(c) = 0$ .
Example: If $3$ is a zero, then $(x - 3)$ is a factor. Conversely, if $(x - 3)$ is a factor, the polynomial divides by $(x - 3)$ perfectly with a remainder of zero.

Evaluating a Function: This is the process of finding an output for a given input.
If $f(x) = x^2$ , evaluating $f(2)$ means replacing the variable $x$ with the input value $2$ , resulting in $2^2 = 4$ .
In the context of the Factor Theorem, evaluating $P(c)$ is used to verify if a remainder is zero. One does not plug $P(c)$ into the divisor $(x-c)$ , but into the original polynomial $P(x)$ .

Degree of a Polynomial ( $n$ ): The highest power of the variable in the polynomial.
Maximum Number of Zeros: A polynomial of degree $n$ can have at most $n$ zeros (and thus at most $n$ factors).
Maximum Number of Turning Points: A polynomial of degree $n$ has at most $n - 1$ turning points.
- Example: A quadratic equation (degree $2$ ) has a maximum of $2$ zeros and $1$ turning point.
- Example: A cubic equation (degree $3$ ) has a maximum of $3$ zeros and $2$ turning points.

Given $P(x) = x^3 - 0x^2 - 7x + 6$ (implied coefficients),
Step 1: Verify a Root. To show that $1$ is a zero, evaluate $P(1)$ . If $P(1) = 0$ , then $(x - 1)$ is a factor.
Step 2: Find Other Factors. Use synthetic division to divide $P(x)$ by the verified factor $(x - 1)$ .
- Coefficients: $[1, 0, -7, 6]$ .
- Synthetic process using root $1$ results in a quotient of $x^2 + x - 6$ .
Step 3: Factor the Quotient. Factor the resulting quadratic expression $x^2 + x - 6$ .
- Find two numbers that multiply to $-6$ and add to $1$ . These are $3$ and $-2$ .
- The factors are $(x + 3)$ and $(x - 2)$ .
Step 4: Combine All Factors. The completely factored form is $P(x) = (x - 1)(x + 3)(x - 2)$ .

To create a polynomial given a set of zeros, convert each zero into its corresponding binomial factor and multiply them.
Example: Construct a polynomial with zeros $-3$ , $0$ , $1$ , and $5$ .
- Zero $-3 \rightarrow$ factor $(x + 3)$ .
- Zero $0 \rightarrow$ factor $(x)$ .
- Zero $1 \rightarrow$ factor $(x - 1)$ .
- Zero $5 \rightarrow$ factor $(x - 5)$ .
Polynomial Expression: $P(x) = x(x + 3)(x - 1)(x - 5)$ .
The degree of this polynomial is $4$ because there are four unique zeros.

Homework Formatting: Students must send photos of homework horizontally rather than vertically to avoid rotation issues. Ensure the photo is not cut off; approximately a quarter of some previous submissions was missing.
Notes Quality: Effective study notes must be recorded after the class when ideas are fresh. Simply writing two or three sentences is insufficient. Students should include every point discussed through interactive questioning.
Clarification for Ethan: In the long division homework, every subtraction step must be shown. One cannot skip the intermediate subtraction of terms and only write the final remainder.
Clarification for Conrad: The remainder discovered through calculation is specifically $P(c)$ . P(4) being the remainder is distinct from the value 4 itself. All important conditions for synthetic division (such as the linear form of the divisor) must be documented in notes.
Clarification for Sakhashv: Homework must be submitted regularly.