Learn on PengiOpenstax Intermediate Algebra 2EChapter 6: Factoring

Lesson 6.4: General Strategy for Factoring Polynomials

In this lesson from OpenStax Intermediate Algebra 2E, students learn how to apply a general step-by-step strategy for factoring polynomials completely, choosing among methods such as greatest common factor, difference of squares, sum or difference of cubes, trinomial factoring, and grouping based on the structure of the polynomial. The lesson emphasizes recognizing whether a polynomial is a binomial, trinomial, or has more than three terms before selecting the correct technique. It is part of Chapter 6 and is appropriate for intermediate algebra students at the high school or early college level.

Section 1

πŸ“˜ General Strategy for Factoring Polynomials

New Concept

This lesson ties everything together! You'll learn a master strategy to decide which factoring method to use for any polynomial, ensuring you can break it down completely and confidently, from finding the GCF to applying special patterns.

What’s next

Next, you'll see this strategy in action through worked examples and then test your skills on a series of challenge problems.

Section 2

General Strategy for Factoring Polynomials

Property

Step 1. Is there a greatest common factor? Factor it out.

Step 2. Is the polynomial a binomial, trinomial, or are there more than three terms?

  • If it is a binomial: Is it a sum of squares (prime), sum of cubes, difference of squares, or difference of cubes?
  • If it is a trinomial: Is it of the form x2+bx+cx^2 + bx + c or ax2+bx+cax^2 + bx + c? Check for perfect square patterns.
  • If it has more than three terms: Use the grouping method.

Step 3. Check. Is it factored completely? Do the factors multiply back to the original polynomial? A polynomial is completely factored if, other than monomials, its factors are prime.

Section 3

Factoring Special Binomials

Property

  • Difference of Squares: a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a-b)(a+b)
  • Sum of Squares: Sums of squares do not factor.
  • Sum of Cubes: a3+b3=(a+b)(a2βˆ’ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2)
  • Difference of Cubes: a3βˆ’b3=(aβˆ’b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Examples

  • To factor the difference of squares 9x2βˆ’499x^2 - 49, recognize it as (3x)2βˆ’72(3x)^2 - 7^2. Applying the formula gives (3xβˆ’7)(3x+7)(3x-7)(3x+7).
  • To factor the sum of cubes 8a3+27b38a^3 + 27b^3, view it as (2a)3+(3b)3(2a)^3 + (3b)^3. The formula yields (2a+3b)(4a2βˆ’6ab+9b2)(2a+3b)(4a^2 - 6ab + 9b^2).

Section 4

Factoring Trinomials

Property

  • Trinomials of the form x2+bx+cx^2 + bx + c: Find two integers that multiply to cc and add to bb, then 'Undo FOIL'.
  • Trinomials of the form ax2+bx+cax^2 + bx + c: Use the trial and error or 'ac' method. If aa and cc are perfect squares, check if it fits a perfect square trinomial pattern.
  • Perfect Square Trinomials:
a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2
a2βˆ’2ab+b2=(aβˆ’b)2a^2 - 2ab + b^2 = (a-b)^2

Examples

  • To factor x2βˆ’2xβˆ’15x^2 - 2x - 15, find two numbers that multiply to βˆ’15-15 and add to βˆ’2-2. These are βˆ’5-5 and 33, so the factors are (xβˆ’5)(x+3)(x-5)(x+3).
  • To factor the perfect square trinomial 9x2+12x+49x^2 + 12x + 4, notice it fits the pattern (3x)2+2(3x)(2)+22(3x)^2 + 2(3x)(2) + 2^2. This factors directly to (3x+2)2(3x+2)^2.

Section 5

Factoring by Grouping

Property

If a polynomial has more than three terms, use the grouping method.

Step 1. Group the polynomial into pairs of terms.

Step 2. Factor out the GCF from each pair.

Section 6

Factoring Completely

Property

A polynomial is completely factored if, other than monomials, its factors are prime. To factor completely:

Step 1. Always factor out the Greatest Common Factor (GCF) first.

Step 2. After factoring, examine each new factor to see if it can be factored further. For example, a difference of squares might appear after the initial step.

Section 7

Grouping a Trinomial and a Term

Property

When a four-term polynomial cannot be factored by pairing, try grouping three terms that form a perfect square trinomial.

Step 1. Identify and factor the perfect square trinomial, resulting in the form (aΒ±b)2(a Β± b)^2.

Step 2. The expression becomes a difference of two squares: (aΒ±b)2βˆ’c2(a Β± b)^2 - c^2.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 6: Factoring

  1. Lesson 1

    Lesson 6.1: Greatest Common Factor and Factor by Grouping

  2. Lesson 2

    Lesson 6.2: Factor Trinomials

  3. Lesson 3

    Lesson 6.3: Factor Special Products

  4. Lesson 4Current

    Lesson 6.4: General Strategy for Factoring Polynomials

  5. Lesson 5

    Lesson 6.5: Polynomial Equations

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ General Strategy for Factoring Polynomials

New Concept

This lesson ties everything together! You'll learn a master strategy to decide which factoring method to use for any polynomial, ensuring you can break it down completely and confidently, from finding the GCF to applying special patterns.

What’s next

Next, you'll see this strategy in action through worked examples and then test your skills on a series of challenge problems.

Section 2

General Strategy for Factoring Polynomials

Property

Step 1. Is there a greatest common factor? Factor it out.

Step 2. Is the polynomial a binomial, trinomial, or are there more than three terms?

  • If it is a binomial: Is it a sum of squares (prime), sum of cubes, difference of squares, or difference of cubes?
  • If it is a trinomial: Is it of the form x2+bx+cx^2 + bx + c or ax2+bx+cax^2 + bx + c? Check for perfect square patterns.
  • If it has more than three terms: Use the grouping method.

Step 3. Check. Is it factored completely? Do the factors multiply back to the original polynomial? A polynomial is completely factored if, other than monomials, its factors are prime.

Section 3

Factoring Special Binomials

Property

  • Difference of Squares: a2βˆ’b2=(aβˆ’b)(a+b)a^2 - b^2 = (a-b)(a+b)
  • Sum of Squares: Sums of squares do not factor.
  • Sum of Cubes: a3+b3=(a+b)(a2βˆ’ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2)
  • Difference of Cubes: a3βˆ’b3=(aβˆ’b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Examples

  • To factor the difference of squares 9x2βˆ’499x^2 - 49, recognize it as (3x)2βˆ’72(3x)^2 - 7^2. Applying the formula gives (3xβˆ’7)(3x+7)(3x-7)(3x+7).
  • To factor the sum of cubes 8a3+27b38a^3 + 27b^3, view it as (2a)3+(3b)3(2a)^3 + (3b)^3. The formula yields (2a+3b)(4a2βˆ’6ab+9b2)(2a+3b)(4a^2 - 6ab + 9b^2).

Section 4

Factoring Trinomials

Property

  • Trinomials of the form x2+bx+cx^2 + bx + c: Find two integers that multiply to cc and add to bb, then 'Undo FOIL'.
  • Trinomials of the form ax2+bx+cax^2 + bx + c: Use the trial and error or 'ac' method. If aa and cc are perfect squares, check if it fits a perfect square trinomial pattern.
  • Perfect Square Trinomials:
a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2
a2βˆ’2ab+b2=(aβˆ’b)2a^2 - 2ab + b^2 = (a-b)^2

Examples

  • To factor x2βˆ’2xβˆ’15x^2 - 2x - 15, find two numbers that multiply to βˆ’15-15 and add to βˆ’2-2. These are βˆ’5-5 and 33, so the factors are (xβˆ’5)(x+3)(x-5)(x+3).
  • To factor the perfect square trinomial 9x2+12x+49x^2 + 12x + 4, notice it fits the pattern (3x)2+2(3x)(2)+22(3x)^2 + 2(3x)(2) + 2^2. This factors directly to (3x+2)2(3x+2)^2.

Section 5

Factoring by Grouping

Property

If a polynomial has more than three terms, use the grouping method.

Step 1. Group the polynomial into pairs of terms.

Step 2. Factor out the GCF from each pair.

Section 6

Factoring Completely

Property

A polynomial is completely factored if, other than monomials, its factors are prime. To factor completely:

Step 1. Always factor out the Greatest Common Factor (GCF) first.

Step 2. After factoring, examine each new factor to see if it can be factored further. For example, a difference of squares might appear after the initial step.

Section 7

Grouping a Trinomial and a Term

Property

When a four-term polynomial cannot be factored by pairing, try grouping three terms that form a perfect square trinomial.

Step 1. Identify and factor the perfect square trinomial, resulting in the form (aΒ±b)2(a Β± b)^2.

Step 2. The expression becomes a difference of two squares: (aΒ±b)2βˆ’c2(a Β± b)^2 - c^2.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 6: Factoring

  1. Lesson 1

    Lesson 6.1: Greatest Common Factor and Factor by Grouping

  2. Lesson 2

    Lesson 6.2: Factor Trinomials

  3. Lesson 3

    Lesson 6.3: Factor Special Products

  4. Lesson 4Current

    Lesson 6.4: General Strategy for Factoring Polynomials

  5. Lesson 5

    Lesson 6.5: Polynomial Equations