Polynomial Expressions and Equations
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Polynomial Vocabulary
Key Terms:
Polynomial:
An expression with variables and coefficients, involving only addition, subtraction, and multiplication with non-negative integer exponents
Degree:
The highest power of the variable in the polynomial
Leading Coefficient:
The coefficient of the term with the highest degree
Constant Term:
The term without a variable (degree 0)
Standard Form:
Terms arranged in descending order of degree
Types by Degree:
• Constant (degree 0): 5
• Linear (degree 1): 2x + 3
• Quadratic (degree 2): \( x^2 + 4x - 5 \)
• Cubic (degree 3): \( x^3 - 2x^2 + x - 1 \)
• Quartic (degree 4): \( x^4 + x^2 + 1 \)
2. Divide Polynomials Using Long Division
Steps:
1. Arrange both polynomials in descending order of degree
2. Insert 0 coefficients for any missing degrees
3. Divide the first term of dividend by first term of divisor
4. Multiply divisor by the result and subtract from dividend
5. Bring down the next term and repeat
6. Continue until degree of remainder < degree of divisor
Example:
Divide: \( (2x^3 + 3x^2 - 5x + 4) \div (x + 2) \)
Result: Quotient = \( 2x^2 - x - 3 \), Remainder = 10
\[ \frac{2x^3 + 3x^2 - 5x + 4}{x + 2} = 2x^2 - x - 3 + \frac{10}{x + 2} \]
3. Divide Polynomials Using Synthetic Division
Requirements:
Synthetic division ONLY works when dividing by a linear expression in the form \( x - c \)
Steps:
1. Write coefficients of dividend in descending order (include 0 for missing terms)
2. Write the value of c (from \( x - c \)) to the left
3. Bring down the first coefficient
4. Multiply by c, place result under next coefficient
5. Add the column and repeat
6. Last number is the remainder; others are quotient coefficients
Example:
Divide: \( (x^3 - 6x^2 + 11x - 6) \div (x - 2) \)
Use c = 2 (from x - 2)
Coefficients: 1, -6, 11, -6
After synthetic division: 1, -4, 3 | 0
Quotient: \( x^2 - 4x + 3 \), Remainder: 0
4. Evaluate Polynomials Using Synthetic Division (Remainder Theorem)
Remainder Theorem:
When polynomial \( P(x) \) is divided by \( (x - c) \), the remainder is \( P(c) \)
\[ P(x) = (x - c) \cdot Q(x) + R, \quad \text{where } R = P(c) \]
Use:
To quickly evaluate \( P(c) \), use synthetic division with c. The remainder is the answer!
Example:
Evaluate \( P(3) \) for \( P(x) = 2x^3 - 5x^2 + 3x - 4 \)
Use synthetic division with c = 3
Coefficients: 2, -5, 3, -4
The remainder is 20
Therefore: \( P(3) = 20 \)
5. Factor Sums and Differences of Cubes
Formulas:
Sum of Cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Difference of Cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Memory Tip (SOAP):
• S = Same sign as the operation
• O = Opposite sign in the trinomial
• AP = Always Positive (last term)
Examples:
Factor: \( x^3 + 8 \)
\( a = x, b = 2 \) (since \( 2^3 = 8 \))
\( x^3 + 8 = (x + 2)(x^2 - 2x + 4) \)
Factor: \( 27x^3 - 64 \)
\( a = 3x, b = 4 \) (since \( (3x)^3 = 27x^3 \) and \( 4^3 = 64 \))
\( 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) \)
6. Solve Equations with Sums and Differences of Cubes
Method:
1. Factor using sum or difference of cubes formula
2. Set each factor equal to zero
3. Solve linear factor (always gives one real solution)
4. Solve quadratic factor (may give 0, 1, or 2 more solutions)
Example:
Solve: \( x^3 - 27 = 0 \)
Factor: \( (x - 3)(x^2 + 3x + 9) = 0 \)
From \( x - 3 = 0 \): \( x = 3 \)
From \( x^2 + 3x + 9 = 0 \): Use quadratic formula
Discriminant: \( 9 - 36 = -27 < 0 \) (no real solutions)
Solution: x = 3 (only real solution)
7. Factor Using a Quadratic Pattern
Method:
When an expression looks like a quadratic but with higher powers, use substitution
Example Pattern:
• For \( x^4 + 5x^2 + 6 \), let \( u = x^2 \)
• Then: \( u^2 + 5u + 6 \)
• Factor: \( (u + 2)(u + 3) \)
• Substitute back: \( (x^2 + 2)(x^2 + 3) \)
Example:
Factor: \( x^4 - 13x^2 + 36 \)
Let \( u = x^2 \), so \( u^2 = x^4 \)
\( u^2 - 13u + 36 = (u - 4)(u - 9) \)
Substitute back: \( (x^2 - 4)(x^2 - 9) \)
Factor further: \( (x - 2)(x + 2)(x - 3)(x + 3) \)
Factored form: \( (x - 2)(x + 2)(x - 3)(x + 3) \)
8. Solve Equations Using a Quadratic Pattern
Example:
Solve: \( x^4 - 5x^2 + 4 = 0 \)
Let \( u = x^2 \): \( u^2 - 5u + 4 = 0 \)
Factor: \( (u - 1)(u - 4) = 0 \)
\( u = 1 \) or \( u = 4 \)
Substitute back:
\( x^2 = 1 \) → \( x = \pm 1 \)
\( x^2 = 4 \) → \( x = \pm 2 \)
Solutions: x = -2, -1, 1, 2
9. Pascal's Triangle
Structure:
A triangular array where each number is the sum of the two numbers above it
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Properties:
• Row 0 starts with 1
• Each row starts and ends with 1
• Row n has (n+1) numbers
• Numbers are binomial coefficients \( \binom{n}{k} \)
10. Pascal's Triangle and the Binomial Theorem
Row n of Pascal's triangle gives the coefficients for \( (a + b)^n \)
Examples:
\( (a + b)^2 = 1a^2 + 2ab + 1b^2 \) (Row 2: 1, 2, 1)
\( (a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3 \) (Row 3: 1, 3, 3, 1)
\( (a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4 \) (Row 4: 1, 4, 6, 4, 1)
11. Binomial Theorem I
The Binomial Theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Expanded Form:
\[ (a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n \]
Binomial Coefficient:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Example:
Expand: \( (x + 2)^4 \)
Using coefficients from Row 4: 1, 4, 6, 4, 1
\( = 1x^4(2)^0 + 4x^3(2)^1 + 6x^2(2)^2 + 4x^1(2)^3 + 1x^0(2)^4 \)
\( = x^4 + 8x^3 + 24x^2 + 32x + 16 \)
12. Binomial Theorem II (Finding Specific Terms)
Finding the (k+1)th Term:
\[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]
This formula finds a specific term without expanding the entire binomial
Example:
Find the 5th term in the expansion of \( (2x - 3)^7 \)
5th term means k = 4 (since k+1 = 5)
n = 7, a = 2x, b = -3
\( T_5 = \binom{7}{4}(2x)^{7-4}(-3)^4 \)
\( = 35(2x)^3(81) \)
\( = 35(8x^3)(81) \)
\( = 22{,}680x^3 \)
13. Quick Reference Summary
Key Formulas:
Sum of Cubes: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
Difference of Cubes: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
Remainder Theorem: When P(x) is divided by (x - c), remainder = P(c)
Binomial Theorem: \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
Binomial Coefficient: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Specific Term: \( T_{k+1} = \binom{n}{k} a^{n-k} b^k \)
📚 Study Tips
✓ Synthetic division only works for divisors in form (x - c)
✓ Use SOAP to remember sum/difference of cubes formulas
✓ Remainder Theorem provides quick polynomial evaluation
✓ Pascal's triangle row n gives coefficients for (a+b)ⁿ
✓ For quadratic patterns, substitute to simplify complex expressions