NUM8ERS Tutoring Center Dubai logo – you can count on us
Book Now
Sign In

AP Precalculus: Polynomial Expressions & Equations

Master polynomial division, factoring techniques, and the binomial theorem for AP exam success

📚 Vocabulary ➗ Division Methods 🧊 Cubes Formulas 📐 Binomial Theorem

📚 Understanding Polynomials

Polynomials are fundamental algebraic expressions consisting of terms with non-negative integer exponents. This guide covers essential vocabulary, division techniques, factoring formulas, and the powerful binomial theorem — all crucial for AP Precalculus success.

1 Polynomial Vocabulary

A polynomial is an expression of the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\) where all exponents are non-negative integers and coefficients are real numbers.

Essential Terms

Degree
Highest power of \(x\) with non-zero coefficient
Leading Coefficient
Coefficient of the highest degree term (\(a_n\))
Constant Term
The term with no variable (\(a_0\))
Leading Term
The term with the highest power (\(a_nx^n\))

Classification by Number of Terms

Monomial
1 term (e.g., \(5x^3\))
Binomial
2 terms (e.g., \(x^2 + 3\))
Trinomial
3 terms (e.g., \(x^2 + 2x + 1\))

Classification by Degree

Constant
Degree 0 (e.g., \(7\))
Linear
Degree 1 (e.g., \(3x + 2\))
Quadratic
Degree 2 (e.g., \(x^2 - 4\))
Cubic
Degree 3 (e.g., \(x^3 + x\))
Quartic
Degree 4 (e.g., \(x^4 - 1\))
Quintic
Degree 5 (e.g., \(x^5 + 2x^3\))
📌 Example

Identify the parts of: \(f(x) = 4x^5 - 3x^3 + 2x - 7\)

Degree: 5 (quintic)

Leading coefficient: 4

Leading term: \(4x^5\)

Constant term: \(-7\)

Number of terms: 4 (polynomial)

2 Polynomial Long Division

Polynomial long division divides a polynomial by another polynomial, similar to numerical long division. It works for any divisor, not just linear factors.

Division Algorithm \[f(x) = d(x) \cdot q(x) + r(x)\] Dividend = Divisor × Quotient + Remainder

Steps for Long Division

  1. Arrange both polynomials in descending order by degree, inserting 0 coefficients for missing terms
  2. Divide the leading term of the dividend by the leading term of the divisor
  3. Multiply the entire divisor by this result and write below
  4. Subtract to get a new polynomial
  5. Repeat steps 2-4 until the degree of the remainder is less than the degree of the divisor
📌 Example

Divide: \((x^3 + 4x^2 - 5x + 2) \div (x - 1)\)

Process:

• \(x^3 \div x = x^2\) → multiply \((x-1)\) by \(x^2\) → subtract

• \(5x^2 \div x = 5x\) → multiply \((x-1)\) by \(5x\) → subtract

• \(0x \div x = 0\) → multiply \((x-1)\) by \(0\) → subtract

Result: Quotient = \(x^2 + 5x + 0\), Remainder = \(2\)

\(x^3 + 4x^2 - 5x + 2 = (x-1)(x^2 + 5x) + 2\)

💡 Pro Tip

Always insert placeholder terms with coefficient 0 for missing powers. For example, when dividing \(x^3 + 1\), write it as \(x^3 + 0x^2 + 0x + 1\).

3 Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear factor of the form \((x - c)\). It uses only coefficients, making it faster than long division.

When to Use Synthetic Division

  • The divisor must be linear: \((x - c)\) or \((x + c)\)
  • Use \(c\) (the root) in the synthetic division — if dividing by \((x - 3)\), use \(3\); if by \((x + 2)\), use \(-2\)
  • Much faster than long division for linear divisors

Steps for Synthetic Division

  1. Write the value \(c\) to the left (from \(x - c\))
  2. Write all coefficients of the dividend in order (use 0 for missing terms)
  3. Bring down the first coefficient
  4. Multiply by \(c\) and add to the next coefficient
  5. Repeat multiply-add until done
  6. Read the result: last number is remainder, others are quotient coefficients
📌 Example

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

Setup: Use \(c = 2\), coefficients: \(2, -5, 3, -7\)

Process:

• Bring down 2

• \(2 \times 2 = 4\); \(-5 + 4 = -1\)

• \(2 \times (-1) = -2\); \(3 + (-2) = 1\)

• \(2 \times 1 = 2\); \(-7 + 2 = -5\)

Result: Quotient = \(2x^2 - x + 1\), Remainder = \(-5\)

⚠️ Common Mistake

When dividing by \((x + 3)\), use \(c = -3\), not \(+3\). Remember: the divisor is \((x - c)\), so \((x + 3) = (x - (-3))\).

4 The Remainder Theorem

The Remainder Theorem states that when a polynomial \(p(x)\) is divided by \((x - c)\), the remainder equals \(p(c)\). This provides a quick way to evaluate polynomials!

Remainder Theorem \[p(c) = \text{remainder when } p(x) \text{ is divided by } (x - c)\]

Applications

  • Evaluate polynomials quickly: Use synthetic division to find \(p(c)\) without substituting
  • Check if \(c\) is a root: If remainder = 0, then \((x - c)\) is a factor and \(c\) is a zero of \(p(x)\)
  • Factor Theorem: \((x - c)\) is a factor of \(p(x)\) if and only if \(p(c) = 0\)
📌 Example

Evaluate: \(p(3)\) where \(p(x) = x^3 - 2x^2 + x - 6\)

Method: Use synthetic division with \(c = 3\)

Coefficients: \(1, -2, 1, -6\)

• Bring down 1

• \(3 \times 1 = 3\); \(-2 + 3 = 1\)

• \(3 \times 1 = 3\); \(1 + 3 = 4\)

• \(3 \times 4 = 12\); \(-6 + 12 = 6\)

Result: \(p(3) = 6\)

💡 When to Use This

The Remainder Theorem is especially useful when you need to evaluate a polynomial at multiple values or when the number has many decimal places — synthetic division is faster than direct substitution.

5 Sum & Difference of Cubes

These special factoring formulas allow you to factor expressions that are the sum or difference of two perfect cubes. Memorize these patterns — they appear frequently on the AP exam!

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 Trick: SOAP

Same sign as original, Opposite sign, Always Positive
For \(a^3 + b^3\): \((a \mathbf{+} b)(a^2 \mathbf{-} ab \mathbf{+} b^2)\) → Same, Opposite, Always Positive

Common Perfect Cubes to Recognize

  • \(1^3 = 1\), \(2^3 = 8\), \(3^3 = 27\), \(4^3 = 64\), \(5^3 = 125\)
  • \(6^3 = 216\), \(7^3 = 343\), \(8^3 = 512\), \(9^3 = 729\), \(10^3 = 1000\)
📌 Examples

Factor: \(x^3 + 27\)

\(= x^3 + 3^3 = (x + 3)(x^2 - 3x + 9)\)

Factor: \(8x^3 - 125\)

\(= (2x)^3 - 5^3 = (2x - 5)(4x^2 + 10x + 25)\)

⚠️ Don't Confuse with Perfect Squares

The trinomial \(a^2 - ab + b^2\) does NOT factor further (it's not a perfect square). The middle term has coefficient 1, not 2.

6 Quadratic Pattern Factoring

Some higher-degree polynomials can be factored using a substitution that transforms them into quadratic form. This technique works when the polynomial has the pattern \(ax^{2n} + bx^n + c\).

Quadratic Substitution For \(ax^{2n} + bx^n + c\), let \(u = x^n\)
Then factor: \(au^2 + bu + c\)

Steps for Quadratic Pattern Factoring

  1. Identify the pattern: Check if exponents are in ratio 2:1:0
  2. Make substitution: Let \(u = x^n\) where \(n\) is the middle exponent
  3. Factor as quadratic: Factor \(au^2 + bu + c\)
  4. Substitute back: Replace \(u\) with \(x^n\)
  5. Factor further if possible
📌 Example 1

Factor: \(x^4 - 5x^2 + 4\)

Step 1: Pattern is \(x^4, x^2, x^0\) — ratio 4:2:0 ✓

Step 2: Let \(u = x^2\): \(u^2 - 5u + 4\)

Step 3: Factor: \((u - 1)(u - 4)\)

Step 4: Substitute back: \((x^2 - 1)(x^2 - 4)\)

Step 5: Factor further: \((x+1)(x-1)(x+2)(x-2)\)

📌 Example 2

Factor: \(x^6 + 7x^3 - 8\)

Let \(u = x^3\): \(u^2 + 7u - 8 = (u + 8)(u - 1)\)

Substitute back: \((x^3 + 8)(x^3 - 1)\)

Factor cubes: \((x+2)(x^2-2x+4)(x-1)(x^2+x+1)\)

7 Pascal's Triangle

Pascal's Triangle is a triangular array of numbers where each entry is the sum of the two entries directly above it. Each row gives the coefficients for expanding \((a + b)^n\).

🔺 Pascal's Triangle (First 6 Rows)
n=0:1
n=1:1   1
n=2:1   2   1
n=3:1   3   3   1
n=4:1   4   6   4   1
n=5:1   5   10   10   5   1

Key Properties

  • Row \(n\) contains the coefficients for \((a + b)^n\)
  • Each row starts and ends with 1
  • Each interior number is the sum of the two numbers above it
  • Row \(n\) has \(n + 1\) entries
  • The entries are binomial coefficients: \(\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}\)
📌 Example: Expand \((a + b)^4\)

Row 4 coefficients: 1, 4, 6, 4, 1

\((a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4\)

\(= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)

8 The Binomial Theorem

The Binomial Theorem provides a formula for expanding any power of a binomial \((a + b)^n\) without multiplying it out step by step.

Binomial Theorem \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
Binomial Coefficient Formula \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] Read as "n choose k"

Understanding the Pattern

  • There are \(n + 1\) terms in the expansion
  • Powers of \(a\) decrease from \(n\) to \(0\)
  • Powers of \(b\) increase from \(0\) to \(n\)
  • The sum of exponents in each term equals \(n\)
  • Coefficients come from Pascal's Triangle or the formula
📌 Example: Expand \((2x + 3)^4\)

Using row 4: coefficients 1, 4, 6, 4, 1

\(= 1(2x)^4(3)^0 + 4(2x)^3(3)^1 + 6(2x)^2(3)^2 + 4(2x)^1(3)^3 + 1(2x)^0(3)^4\)

\(= 16x^4 + 4(8x^3)(3) + 6(4x^2)(9) + 4(2x)(27) + 81\)

\(= 16x^4 + 96x^3 + 216x^2 + 216x + 81\)

Finding a Specific Term

The \((k+1)\)th Term \[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]
📌 Example: Find the 4th term of \((x + 2)^6\)

4th term means \(k = 3\) (since \(k+1 = 4\))

\(T_4 = \binom{6}{3} x^{6-3} \cdot 2^3 = \frac{6!}{3!3!} \cdot x^3 \cdot 8 = 20 \cdot 8x^3 = 160x^3\)

💡 Calculator Tip

Most calculators have an nCr function to compute \(\binom{n}{k}\). On TI calculators, use MATH → PRB → nCr.

📋 Quick Reference: Key Formulas

Division Algorithm

\(f(x) = d(x) \cdot q(x) + r(x)\)

Remainder Theorem

\(p(c) = \) remainder when dividing by \((x-c)\)

Sum of Cubes

\(a^3 + b^3 = (a+b)(a^2-ab+b^2)\)

Difference of Cubes

\(a^3 - b^3 = (a-b)(a^2+ab+b^2)\)

Binomial Coefficient

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

Binomial Theorem

\((a+b)^n = \sum \binom{n}{k}a^{n-k}b^k\)

Need Help Mastering Polynomial Concepts?

Our expert tutors provide personalized instruction to help you excel in AP Precalculus.

Book Free Consultation