6.3 Factor Special Products

Factoring special products lets you break down certain expressions quickly by recognizing their structure instead of using trial and error. Three patterns show up constantly: perfect square trinomials, differences of squares, and sums/differences of cubes. Once you can spot these forms, factoring becomes much faster.

Factoring Special Products

Perfect Square Trinomial Factoring

A perfect square trinomial comes from squaring a binomial. It takes one of two forms:

• $a^2 + 2ab + b^2 = (a + b)^2$
• $a^2 - 2ab + b^2 = (a - b)^2$

The key feature is that middle term: it's always twice the product of the square roots of the first and last terms. That's how you confirm you're dealing with a perfect square trinomial and not just a regular trinomial.

Steps to factor:

1. Check whether the first and last terms are both perfect squares. Find their square roots ($a$ and $b$).
2. Verify the middle term equals $2ab$. If it doesn't, this pattern doesn't apply.
3. Look at the sign of the middle term:
   • Positive middle term → $(a + b)^2$
   • Negative middle term → $(a - b)^2$

Example: Factor $x^2 + 6x + 9$.

• $x^2$ is a perfect square (root: $x$). $9$ is a perfect square (root: $3$).
• Middle term check: $2(x)(3) = 6x$. That matches.
• Middle term is positive, so the answer is $(x + 3)^2$.

Example: Factor $x^2 - 10x + 25$.

• Roots are $x$ and $5$. Middle term check: $2(x)(5) = 10x$. That matches.
• Middle term is negative, so the answer is $(x - 5)^2$.

This pattern is closely related to completing the square, which you'll use later to solve quadratic equations and rewrite them in vertex form.

Differences of Squares Factoring

A difference of squares is an expression of the form $a^2 - b^2$. It factors as:

$a^2 - b^2 = (a + b)(a - b)$

This works because when you multiply $(a + b)(a - b)$ using FOIL, the middle terms cancel out, leaving only $a^2 - b^2$. Notice there's no "sum of squares" version: $a^2 + b^2$ does not factor over the real numbers.

Steps to factor:

1. Confirm both terms are perfect squares and they're being subtracted.
2. Find the square roots of each term ($a$ and $b$).
3. Write the factored form as $(a + b)(a - b)$.

Example: Factor $x^2 - 16$.

• $x^2$ has root $x$. $16$ has root $4$. Subtraction? Yes.
• Answer: $(x + 4)(x - 4)$.

The two factors $(a + b)$ and $(a - b)$ are called conjugates. You'll see conjugates again when rationalizing denominators and working with complex numbers.

Sums and Differences of Cubes

These patterns handle expressions where two terms are cubed:

• Sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
• Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

A helpful mnemonic: the first factor $(a \pm b)$ uses the same sign as the original expression. In the second factor, the signs follow the pattern opposite, always positive. So for a sum of cubes, you get $+$, then $-$, $+$. For a difference of cubes, you get $-$, then $+$, $+$.

Steps to factor:

1. Confirm both terms are perfect cubes. Find their cube roots ($a$ and $b$).

2. Write the first factor using the same sign as the original: $(a + b)$ or $(a - b)$.

3. Build the second factor using the pattern:

   • Square the first cube root: $a^2$
   • Multiply the two cube roots with the opposite sign: $\mp ab$
   • Square the second cube root: $+ b^2$

Example: Factor $x^3 + 8$.

• Cube roots: $x$ and $2$ (since $2^3 = 8$).
• First factor (same sign): $(x + 2)$.
• Second factor: $x^2 - 2x + 4$.
• Answer: $(x + 2)(x^2 - 2x + 4)$.

Example: Factor $x^3 - 27$.

• Cube roots: $x$ and $3$ (since $3^3 = 27$).
• First factor (same sign): $(x - 3)$.
• Second factor: $x^2 + 3x + 9$.
• Answer: $(x - 3)(x^2 + 3x + 9)$.

Additional Factoring Techniques

Before applying any special product pattern, always factor out the GCF first. For example, $2x^2 - 18$ doesn't look like a difference of squares at first, but pulling out the 2 gives you $2(x^2 - 9) = 2(x + 3)(x - 3)$.

Also watch for expressions that need factoring more than once. Something like $x^4 - 16$ is a difference of squares: $(x^2 + 4)(x^2 - 4)$. But $x^2 - 4$ is itself a difference of squares, so the fully factored form is $(x^2 + 4)(x + 2)(x - 2)$.

When an expression doesn't match any special pattern and regular factoring methods don't work, the quadratic formula can still solve the related equation directly.