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Special products

In factoring, special products are algebraic expressions that follow very recognizable patterns. Knowing these patterns allows you to factor (and expand) quickly without doing all intermediate multiplication.

This chapter focuses on three main families:

You should already be familiar with basic polynomial operations and simple factoring of common factors from the parent chapters.

Square of a binomial

A binomial is an expression with two terms, like $x+3$ or $2a-5$.

When you square a binomial, there is a consistent pattern:

  1. Square of a sum:
    $$(a + b)^2 = a^2 + 2ab + b^2$$
  2. Square of a difference:
    $$(a - b)^2 = a^2 - 2ab + b^2$$

Notice:

Recognizing $(a+b)^2$ and $(a-b)^2$ when factoring

To factor a quadratic expression as a square of a binomial, look for:

Example:

This matches the pattern $a^2 + 2ab + b^2$ with $a=x$, $b=3$, so:
$$x^2 + 6x + 9 = (x + 3)^2$$

This matches $a^2 - 2ab + b^2$ with $a=2y$, $b=3$, so:
$$4y^2 - 12y + 9 = (2y - 3)^2$$

Be careful: not every expression with square first and last terms is a perfect square trinomial. The middle term must fit the $2ab$ pattern.

Example:

Difference of squares

Another very common special product involves the difference of two squares.

Pattern for expanding:
$$(a - b)(a + b) = a^2 - b^2$$

In factoring, we use this in reverse:

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

Key features:

Example:

So:
$$x^2 - 9 = (x - 3)(x + 3)$$

So:
$$25y^2 - 16 = (5y - 4)(5y + 4)$$

Recognizing difference of squares

Look for:

Not a difference of squares:

For example, with $3x^2 - 12$:

First factor out the greatest common factor (GCF):
$$3x^2 - 12 = 3(x^2 - 4)$$

Now $x^2 - 4$ is a difference of squares:
$$x^2 - 4 = (x - 2)(x + 2)$$

So the fully factored form is:
$$3x^2 - 12 = 3(x - 2)(x + 2)$$

Product of a sum and difference

Closely related to difference of squares is the product of a sum and difference:

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

This is actually the same pattern as before, just written in product form.

When factoring:

Example:

Here $a = 2x$, $b = 5$:
$$(2x + 5)(2x - 5) = (2x)^2 - 5^2 = 4x^2 - 25$$

So:
$$4x^2 - 25 = (2x + 5)(2x - 5)$$

Note: $(a + b)(a - b)$ and $(a - b)(a + b)$ are the same up to order; multiplication is commutative.

Recognizing and using special products when factoring

When you factor a polynomial, it is often helpful to check for special product patterns after simpler steps like taking out a common factor:

  1. Look for a common factor and factor it out.
  2. Check what remains:
    • Is it a difference of squares?
    • Is it a perfect square trinomial (square of a binomial)?

Example:

Factor $x^4 - 10x^2 + 25$.

So:
$$x^4 - 10x^2 + 25 = (x^2 - 5)^2$$

You could then check if $x^2 - 5$ can be factored further over the numbers you are using (over real numbers, $x^2 - 5$ is a difference of squares only if you allow $\sqrt{5}$).

Another example:

Factor $2x^3 - 18x$.

Step 1: Factor out the GCF:
$$2x^3 - 18x = 2x(x^2 - 9)$$

Step 2: Recognize difference of squares in $x^2 - 9$:
$$x^2 - 9 = (x - 3)(x + 3)$$

So:
$$2x^3 - 18x = 2x(x - 3)(x + 3)$$

Here, recognizing $x^2 - 9$ as a difference of squares is the special product step.

Summary of patterns

For quick reference:

Being able to spot these patterns makes factoring (and expanding) polynomials faster and more reliable, and they appear frequently in later algebra topics.

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