A polynomial is an algebraic expression consisting of variables and coefficients, using only addition, subtraction, multiplication, and non-negative integer exponents.

\[ P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0 \]

where \( a_n, a_{n-1}, \ldots, a_0 \) are constants (coefficients) and \( n \) is a non-negative integer

• Term: A single part of a polynomial (e.g., \( 3x^2 \))

• Coefficient: The numerical factor of a term (e.g., 3 in \( 3x^2 \))

• Leading Coefficient: The coefficient of the term with the highest degree (e.g., 5 in \( 5x^3 + 2x - 1 \))

• Constant Term: The term without a variable (e.g., -1 in \( 5x^3 + 2x - 1 \))

• Degree: The highest exponent of the variable (e.g., degree 3 in \( 5x^3 + 2x - 1 \))

• Standard Form: Polynomial written with terms in descending order of degree

Rule: Combine like terms

Like terms have the same variable(s) raised to the same power(s)

Steps:

1. Remove parentheses (if any)

2. Group like terms together

3. Add the coefficients of like terms

4. Write result in standard form

Add: \( (3x^2 + 5x - 7) + (2x^2 - 3x + 4) \)

Step 1: Remove parentheses: \( 3x^2 + 5x - 7 + 2x^2 - 3x + 4 \)

Step 2: Group like terms: \( (3x^2 + 2x^2) + (5x - 3x) + (-7 + 4) \)

Step 3: Combine: \( 5x^2 + 2x - 3 \)

Answer: \( 5x^2 + 2x - 3 \)

Rule: Distribute the negative sign, then combine like terms

Steps:

1. Distribute negative sign to each term in second polynomial

2. Remove parentheses

3. Group like terms together

4. Combine coefficients

Subtract: \( (5x^2 - 12x + 1) - (2x^2 - 3x + 7) \)

Step 1: Distribute negative: \( 5x^2 - 12x + 1 - 2x^2 + 3x - 7 \)

Step 2: Group like terms: \( (5x^2 - 2x^2) + (-12x + 3x) + (1 - 7) \)

Step 3: Combine: \( 3x^2 - 9x - 6 \)

Answer: \( 3x^2 - 9x - 6 \)

Multiply each term of the first polynomial by each term of the second polynomial, then combine like terms.

Use the Distributive Property

Example:

\( 3x(2x^2 - 5x + 4) \)

\( = 3x \cdot 2x^2 - 3x \cdot 5x + 3x \cdot 4 \)

\( = 6x^3 - 15x^2 + 12x \)

FOIL: First, Outer, Inner, Last

\( (a + b)(c + d) = ac + ad + bc + bd \)

Example:

\( (2x + 3)(x - 5) \)

F: \( 2x \cdot x = 2x^2 \)

O: \( 2x \cdot (-5) = -10x \)

I: \( 3 \cdot x = 3x \)

L: \( 3 \cdot (-5) = -15 \)

Combine: \( 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15 \)

1. Difference of Squares:

\[ (a + b)(a - b) = a^2 - b^2 \]

Example: \( (x + 4)(x - 4) = x^2 - 16 \)

2. Perfect Square Trinomial:

\[ (a + b)^2 = a^2 + 2ab + b^2 \]

\[ (a - b)^2 = a^2 - 2ab + b^2 \]

Example: \( (x + 3)^2 = x^2 + 6x + 9 \)

Example: \( (x - 5)^2 = x^2 - 10x + 25 \)

Step 1: Arrange both polynomials in descending order

Add placeholder terms with coefficient 0 for missing powers

Step 2: Divide the first term of dividend by first term of divisor

This gives the first term of the quotient

Step 3: Multiply the entire divisor by this term

Step 4: Subtract the result from the dividend

Step 5: Bring down the next term

Step 6: Repeat steps 2-5 until degree of remainder < degree of divisor

Divide: \( (2x^3 + 5x^2 - 3x + 7) \div (x + 2) \)

         2x² + x - 5
    ___________________
x+2 | 2x³ + 5x² - 3x + 7
      2x³ + 4x²
      ___________
            x² - 3x
            x² + 2x
            _______
               -5x + 7
               -5x - 10
               ________
                    17
          

\[ \text{Dividend} = (\text{Divisor})(\text{Quotient}) + \text{Remainder} \]

or

\[ \frac{\text{Dividend}}{\text{Divisor}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}} \]

Synthetic division is a shortcut method that ONLY works when:

• Divisor is linear (degree 1)

• Divisor is in the form \( (x - c) \) where \( c \) is a constant

⚠️ Cannot use for divisors like \( (2x - 3) \) or \( (x^2 + 1) \)

Step 1: Write the value of \( c \) from \( (x - c) \) on the left

If divisor is \( (x + 2) \), write -2; if \( (x - 3) \), write 3

Step 2: Write coefficients of dividend in descending order

Include 0 for missing powers

Step 3: Bring down the first coefficient

Step 4: Multiply by \( c \), write result under next coefficient

Step 5: Add the column, write result below

Step 6: Repeat steps 4-5 until done

Step 7: Last number is remainder, others are quotient coefficients

Divide: \( (2x^3 + 5x^2 - 3x + 7) \div (x + 2) \)

-2 |  2    5   -3    7
   |      -4   -2   10
   |____________________
      2    1   -5   17
          

If a polynomial \( P(x) \) is divided by \( (x - c) \), the remainder is \( P(c) \)

\[ P(c) = \text{Remainder when } P(x) \div (x - c) \]

This means:

We can evaluate \( P(c) \) by using synthetic division and reading the remainder!

Evaluate \( P(3) \) where \( P(x) = x^3 - 4x^2 + 5x - 2 \)

3 |  1   -4    5   -2
  |       3   -3    6
  |___________________
     1   -1    2    4
          

Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it.

                1                 Row 0
              1   1               Row 1
            1   2   1             Row 2
          1   3   3   1           Row 3
        1   4   6   4   1         Row 4
      1   5  10  10   5   1       Row 5
    1   6  15  20  15   6   1     Row 6
        

Pattern Rules:

• Start and end each row with 1

• Each interior number = sum of two numbers above it

• Row \( n \) has \( n + 1 \) numbers

The numbers in Row \( n \) are the binomial coefficients \( \binom{n}{k} \) for \( k = 0, 1, 2, \ldots, n \)

Row 4 example:

\( \binom{4}{0} = 1, \; \binom{4}{1} = 4, \; \binom{4}{2} = 6, \; \binom{4}{3} = 4, \; \binom{4}{4} = 1 \)

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Expanded Form:

\[ (a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n \]

where:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

This is called "n choose k" or the binomial coefficient

Example: Expand \( (x + y)^4 \)

Use Row 4 of Pascal's Triangle: 1, 4, 6, 4, 1

• There are \( n + 1 \) terms in the expansion of \( (a + b)^n \)

• Powers of \( a \) decrease from \( n \) to 0

• Powers of \( b \) increase from 0 to \( n \)

• Sum of exponents in each term equals \( n \)

• Coefficients come from Pascal's Triangle or \( \binom{n}{k} \)

The \( (k+1) \)th term in the expansion of \( (a + b)^n \) is:

\[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]

Example: Find the 3rd term of \( (2x + 3)^5 \)

3rd term means \( k = 2 \) (since \( k+1 = 3 \))

\( T_3 = \binom{5}{2} (2x)^{5-2} (3)^2 \)

\( = 10 \cdot (2x)^3 \cdot 9 \)

\( = 10 \cdot 8x^3 \cdot 9 \)

\( = 720x^3 \)