Factoring Rules

Q: What is the difference between the difference of cubes and the sum of cubes formulas?

The difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ has a minus sign in the binomial factor and all plus signs in the trinomial. The sum of cubes $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ has a plus in the binomial and an alternating pattern (plus, minus, plus) in the trinomial. A helpful mnemonic is "SOAP": Same sign, Opposite sign, Always Positive — referring to the signs in order across both factors.

Q: Can you factor a sum of squares like $a^2 + b^2$?

No, $a^2 + b^2$ cannot be factored into real-number factors. It only factors over the complex numbers as $(a - bi)(a + bi)$, where $i = \sqrt{-1}$. This is a key distinction from the difference of squares $a^2 - b^2$, which always factors over the reals.

Q: When should you use the general rule $a^n - b^n$ instead of a specific formula?

Use the general rule $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$ when $n$ is larger than 3 and no simpler factoring path exists. For even exponents like $n = 4$, you can often factor in stages — for example, treating $a^4 - b^4$ as a difference of squares first, then factoring again. Choosing the most efficient path depends on the specific exponent.

Factoring Rules

Algebra formulas for factoring.

1. x² – (r + s)x + rs = (x – r)(x – s)
2. x² + 2ax + a² = (x + a)² and x² – 2ax + a² = (x – a)²
3. Difference of squares: a² – b² = (a – b)(a + b)
4. Difference of cubes: a³ – b³ = (a – b)(a² + ab + b²)
5. a⁴ – b⁴ = (a – b)(a³ + a²b + ab² + b³) = (a – b) [ a²(a + b) + b²(a + b) ] = (a – b)(a + b)(a² + b²)
	or, more simply, a⁴ – b⁴ = (a² – b²)(a² + b²) = (a – b)(a + b)(a² + b²)
6. a⁵ – b⁵ = (a – b)(a⁴ + a³b + a²b² + ab³ + b⁴)
7. aⁿ – bⁿ = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b² + ··· + ab^{n – 2} + b^{n – 1})
8. Sum of cubes: a³ + b³ = (a + b)(a² – ab + b²)
9. a⁵ + b⁵ = (a + b)(a⁴ – a³b + a²b² – ab³ + b⁴)
10. a⁷ + b⁷ = (a + b)(a⁶ – a⁵b + a⁴b² – a³b³ + a²b⁴ – ab⁵ + b⁶)
11. If n is odd, then aⁿ + bⁿ = (a + b)(a^{n – 1} – a^{n – 2}b + a^{n – 3}b² – ··· + a²b^{n – 3} – ab^{n – 2} + b^{n – 1})
12. Sum of squares: a² + b² = (a – bi)(a + bi) Note: a² + b² does not factor using real numbers.
13.

Key Formula

\begin{aligned} &1.\; x^2 - (r+s)x + rs = (x-r)(x-s) \\ &2.\; x^2 \pm 2ax + a^2 = (x \pm a)^2 \\ &3.\; a^2 - b^2 = (a-b)(a+b) \\ &4.\; a^3 - b^3 = (a-b)(a^2+ab+b^2) \\ &5.\; a^3 + b^3 = (a+b)(a^2-ab+b^2) \\ &6.\; a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}) \end{aligned}

Where:

$a, b$ = Any expressions (numbers, variables, or combinations)
$r, s$ = The roots of the quadratic; the values that make each factor zero
$n$ = A positive integer exponent
$x$ = The variable in the polynomial being factored

Worked Example

Problem: Factor the expression

x^2 - 5x + 6

completely.

Step 1: Identify the form. This is a quadratic

x^2 - (r+s)x + rs

, so you need two numbers

r

and

s

whose sum is 5 and whose product is 6.

r + s = 5 \quad \text{and} \quad rs = 6

Step 2: Find the pair. The numbers 2 and 3 satisfy both conditions:

2 + 3 = 5

and

2 × 3 = 6

r = 2,\; s = 3

Step 3: Apply Rule 1 to write the factored form.

x^2 - 5x + 6 = (x - 2)(x - 3)

Step 4: Verify by expanding:

(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6

. The factoring is correct.

\checkmark

Answer:

x^2 - 5x + 6 = (x - 2)(x - 3)

Another Example

This example uses the difference of cubes rule instead of the basic trinomial rule, and it shows how to handle coefficients other than 1 by rewriting terms as perfect cubes.

Problem: Factor

8x^3 - 27

completely.

Step 1: Recognize that both terms are perfect cubes. Write each term as a cube:

8x^3 = (2x)^3

and

27 = 3^3

8x^3 - 27 = (2x)^3 - 3^3

Step 2: Apply the difference of cubes rule (Rule 4):

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

, with

a = 2x

and

b = 3

a = 2x,\quad b = 3

Step 3: Substitute into the formula. Compute each part of the trinomial factor:

(2x)^2 = 4x^2

(2x)(3) = 6x

, and

3^2 = 9

(2x - 3)\bigl(4x^2 + 6x + 9\bigr)

Step 4: Check that the trinomial

4x^2 + 6x + 9

does not factor further over the reals. Its discriminant is

36 - 144 = -108 < 0

, so it is irreducible.

8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9)

Answer:

8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9)

Frequently Asked Questions

What is the difference between the difference of cubes and the sum of cubes formulas?

The difference of cubes

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

has a minus sign in the binomial factor and all plus signs in the trinomial. The sum of cubes

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

has a plus in the binomial and an alternating pattern (plus, minus, plus) in the trinomial. A helpful mnemonic is "SOAP": Same sign, Opposite sign, Always Positive — referring to the signs in order across both factors.

Can you factor a sum of squares like

a^2 + b^2

No,

a^2 + b^2

cannot be factored into real-number factors. It only factors over the complex numbers as

(a - bi)(a + bi)

, where

i = \sqrt{-1}

. This is a key distinction from the difference of squares

a^2 - b^2

, which always factors over the reals.

When should you use the general rule

a^n - b^n

instead of a specific formula?

Use the general rule

a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})

when

n

is larger than 3 and no simpler factoring path exists. For even exponents like

n = 4

, you can often factor in stages — for example, treating

a^4 - b^4

as a difference of squares first, then factoring again. Choosing the most efficient path depends on the specific exponent.

Factoring vs. Expanding (Distributing)

	Factoring	Expanding (Distributing)
Direction	Breaks a product apart: rewrites a polynomial as a product of factors	Multiplies factors out: rewrites a product as a sum of terms
Example	$x^2 - 9 = (x-3)(x+3)$	$(x-3)(x+3) = x^2 - 9$
When to use	Solving equations, simplifying fractions, finding roots	Simplifying expressions, verifying a factoring result, distributing
Difficulty	Requires pattern recognition; harder because multiple rules may apply	Mechanical process; apply the distributive property systematically

Why It Matters

Factoring rules appear constantly in algebra and precalculus, especially when solving quadratic and higher-degree polynomial equations. You also need them to simplify rational expressions (algebraic fractions) by canceling common factors. In calculus, factoring is used to evaluate limits, integrate rational functions via partial fractions, and simplify derivatives — making fluency with these rules essential long before and well beyond Algebra 2.

Common Mistakes

Mistake: Confusing the signs in the sum vs. difference of cubes trinomial factor. Students often write

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

with a plus on the

ab

term.

Correction: The middle term of the trinomial always has the opposite sign from the binomial factor. For

a^3 + b^3

, the binomial is

(a + b)

, so the trinomial must be

(a^2 - ab + b^2)

. Use the SOAP mnemonic: Same, Opposite, Always Positive.

Mistake: Trying to factor a sum of squares

a^2 + b^2

(a + b)^2

(a + b)(a - b)

Correction: Neither of those is correct. Expanding

(a + b)^2

gives

a^2 + 2ab + b^2

, and

(a + b)(a - b)

gives

a^2 - b^2

. The sum of two squares does not factor over the real numbers. Only the difference of squares

a^2 - b^2

factors as

(a - b)(a + b)

Related Terms

Factor of a Polynomial — Defines what a factor of a polynomial is
Algebra — The branch of math where factoring rules are used
Factor Theorem — Links roots to factors of polynomials
Rational Root Theorem — Identifies possible rational roots to aid factoring
Polynomial Long Division — Divides polynomials when pattern-based factoring fails
Synthetic Division — A shortcut for dividing by linear factors
Real Numbers — The number system over which most factoring is performed
Odd Number — Odd exponents allow sum-of-powers factoring