The Central Idea

Many algebraic expressions can be written in either an expanded form or a factored form.

For example,

\[ (x+2)(x+5) \]

can be expanded to

\[ x^2+7x+10. \]

Conversely,

\[ x^2+7x+10 \]

can be factored as

\[ (x+2)(x+5). \]

These are not different mathematical quantities. They are different representations of the same polynomial.

Expansion emphasizes addition and individual terms. Factoring emphasizes multiplication and common structure.

Expansion

Expansion rewrites a product as an equivalent sum or difference by using the distributive property.

For example,

\[ 3(x+4)=3x+12. \]

For a product of two binomials, distribution is applied more than once:

\[ (x+2)(x+5). \]

Distribute \(x+2\) across the second factor:

\[ x(x+5)+2(x+5). \]

Then distribute again:

\[ x^2+5x+2x+10. \]

Combine like terms:

\[ x^2+7x+10. \]

Therefore,

\[ (x+2)(x+5)=x^2+7x+10. \]

FOIL Is a Shortcut for Distribution

For two binomials, students often learn the mnemonic FOIL: First, Outer, Inner, Last.

For

\[ (a+b)(c+d), \]

FOIL produces

\[ ac+ad+bc+bd. \]

But FOIL is not a separate algebraic law. It is simply a memory aid for repeated distribution.

The distributive property is more general because it also works for expressions with more than two terms.

Expanding Two Binomials

Consider

\[ (x+3)(x-4). \]

Distribute:

\[ x(x-4)+3(x-4). \]

Then

\[ x^2-4x+3x-12. \]

Combine like terms:

\[ x^2-x-12. \]

So

\[ (x+3)(x-4)=x^2-x-12. \]

Expanding with Coefficients

Now consider

\[ (2x+3)(x+4). \]

Distribute:

\[ 2x(x+4)+3(x+4). \]

Then

\[ 2x^2+8x+3x+12. \]

Combine like terms:

\[ 2x^2+11x+12. \]

Therefore,

\[ (2x+3)(x+4)=2x^2+11x+12. \]

Expanding More Than Two Factors

Expressions may contain several factors.

For example,

\[ (x+1)(x+2)(x+3). \]

One approach is to multiply two factors first:

\[ (x+1)(x+2)=x^2+3x+2. \]

Then multiply by the remaining factor:

\[ (x^2+3x+2)(x+3). \]

Distribute:

\[ x^3+3x^2+3x^2+9x+2x+6. \]

Combine like terms:

\[ x^3+6x^2+11x+6. \]

The associative property of multiplication lets us choose which factors to multiply first.

Factoring

Factoring rewrites an expression as a product of simpler expressions.

Factoring is closely related to multiplication in arithmetic.

For example,

\[ 12=3\cdot4=2\cdot6=2^2\cdot3. \]

Likewise, a polynomial can often be decomposed into factors:

\[ x^2+7x+10=(x+2)(x+5). \]

The factored form reveals multiplicative structure that may be hidden in expanded form.

Always Check for a Common Factor First

The first factoring question should usually be:

Do all terms share a common factor?

Consider

\[ 6x^2+9x. \]

Both terms contain \(3x\), so

\[ 6x^2+9x=3x(2x+3). \]

This is the reverse of distribution:

\[ 3x(2x+3)=6x^2+9x. \]

Extracting a greatest common factor often makes additional structure easier to see.

Greatest Common Factor

Consider

\[ 12x^3y+18x^2y^2. \]

The numerical coefficients share a greatest common factor of \(6\).

Both terms also contain at least

\[ x^2y. \]

So a greatest common factor is

\[ 6x^2y. \]

Factor it out:

\[ 12x^3y+18x^2y^2 = 6x^2y(2x+3y). \]

Factoring a Simple Trinomial

Consider

\[ x^2+7x+10. \]

We look for two numbers whose product is \(10\) and whose sum is \(7\).

The numbers are \(2\) and \(5\), so

\[ x^2+7x+10=(x+2)(x+5). \]

Why does this work?

Expanding the right side gives

\[ x^2+(2+5)x+(2)(5), \]

which is

\[ x^2+7x+10. \]

The middle coefficient comes from the sum of the two constants, while the constant term comes from their product.

Factoring When the Constant Is Negative

Consider

\[ x^2+x-12. \]

We need two numbers whose product is \(-12\) and whose sum is \(1\).

The numbers are \(4\) and \(-3\).

Therefore,

\[ x^2+x-12=(x+4)(x-3). \]

Checking by expansion:

\[ (x+4)(x-3)=x^2+x-12. \]

Factoring When the Leading Coefficient Is Not 1

Consider

\[ 2x^2+7x+3. \]

A possible factorization is

\[ (2x+1)(x+3). \]

Expand to verify:

\[ 2x^2+6x+x+3 \]

which becomes

\[ 2x^2+7x+3. \]

For more complicated trinomials, several systematic factoring methods are possible. The essential goal remains the same: find factors whose expansion reproduces the original polynomial.

Factoring by Grouping

Some expressions can be factored by grouping terms into smaller collections.

Consider

\[ ax+ay+bx+by. \]

Group the first two terms and the last two terms:

\[ (ax+ay)+(bx+by). \]

Factor each group:

\[ a(x+y)+b(x+y). \]

Now \(x+y\) is a common factor:

\[ (a+b)(x+y). \]

Thus,

\[ ax+ay+bx+by=(a+b)(x+y). \]

A Numerical-Looking Grouping Example

Factor

\[ x^3+3x^2+2x+6. \]

Group:

\[ (x^3+3x^2)+(2x+6). \]

Factor each group:

\[ x^2(x+3)+2(x+3). \]

Collect the repeated factor:

\[ (x^2+2)(x+3). \]

Therefore,

\[ x^3+3x^2+2x+6=(x^2+2)(x+3). \]

Difference of Squares

One of the most useful factoring patterns is the difference of squares:

\[ a^2-b^2=(a-b)(a+b). \]

For example,

\[ x^2-25 \]

is

\[ x^2-5^2, \]

so

\[ x^2-25=(x-5)(x+5). \]

Checking by expansion:

\[ (x-5)(x+5)=x^2-25. \]

The middle terms cancel.

Perfect Square Trinomials

Two important expansion patterns are

\[ (a+b)^2=a^2+2ab+b^2 \]

and

\[ (a-b)^2=a^2-2ab+b^2. \]

These can also be read backward as factoring patterns.

For example,

\[ x^2+6x+9 \]

matches

\[ x^2+2(x)(3)+3^2. \]

Therefore,

\[ x^2+6x+9=(x+3)^2. \]

Likewise,

\[ x^2-10x+25=(x-5)^2. \]

Sum and Difference of Cubes

Higher-power patterns can also be useful.

The sum of cubes factors as

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

The difference of cubes factors as

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

For example,

\[ x^3-8 \]

is

\[ x^3-2^3, \]

so

\[ x^3-8=(x-2)(x^2+2x+4). \]

These patterns become more useful as polynomial work becomes more advanced.

Recognizing Structure Before Calculating

Factoring becomes easier when we first classify the expression.

For example:

Expression Useful Observation
\(8x+12\) Common factor
\(x^2+9x+20\) Quadratic trinomial
\(x^2-49\) Difference of squares
\(x^2+10x+25\) Perfect square trinomial
\(x^3+2x^2+3x+6\) Possible grouping

The structure suggests which transformation is likely to be useful.

Factoring Can Continue in Stages

Sometimes one factoring step is not enough.

Consider

\[ 2x^2-18. \]

First factor out the common factor \(2\):

\[ 2(x^2-9). \]

Then recognize a difference of squares:

\[ 2(x-3)(x+3). \]

A complete factorization usually continues until no obvious factorization remains over the number system being used.

Factoring Depends on the Number System

An expression may be factorable over one number system but not another.

For example,

\[ x^2+1 \]

does not factor into linear factors over the real numbers.

But over the complex numbers,

\[ x^2+1=(x-i)(x+i). \]

So the phrase “fully factored” always depends on the allowed kinds of factors.

At this stage, most factoring problems will use integers or real-number coefficients unless stated otherwise.

Why Factored Form Is Useful

Factored form often makes zeros visible.

Consider

\[ x^2-5x+6=0. \]

Factor:

\[ (x-2)(x-3)=0. \]

A product equals zero only if at least one factor equals zero:

\[ x-2=0 \]

or

\[ x-3=0. \]

Therefore,

\[ x=2 \]

or

\[ x=3. \]

The expanded form hides these solutions; the factored form exposes them.

Why Expanded Form Is Useful

Factored form is not always preferable.

Consider

\[ (x+4)(x-1). \]

Expanded form is

\[ x^2+3x-4. \]

This form makes the polynomial's individual coefficients visible and may be more convenient for addition, subtraction, coefficient comparison, or certain computational methods.

The best form depends on the task.

Expansion and Factoring Are Reverse Directions

At a basic level, expansion and factoring move through the same equivalence in opposite directions:

\[ (x+2)(x+5) \longleftrightarrow x^2+7x+10. \]

Expansion moves from product structure toward additive structure.

Factoring moves from additive structure toward product structure.

Because both forms are equal for every value of \(x\), either may replace the other whenever useful.

Checking a Factorization

The safest general check for a proposed factorization is simple:

Expand it.

Suppose we believe

\[ x^2-2x-15=(x-5)(x+3). \]

Expand the right side:

\[ x^2+3x-5x-15. \]

Combine:

\[ x^2-2x-15. \]

The expansion matches the original expression, so the factorization is correct.

Factoring Is Not Canceling

Factoring rewrites an expression as a product. It does not by itself remove anything.

For example,

\[ x^2-9=(x-3)(x+3). \]

No factor disappears.

Cancellation occurs only within a quotient when the same nonzero factor appears in both numerator and denominator.

For example,

\[ \frac{(x-3)(x+3)}{x-3}=x+3 \]

provided that

\[ x\ne3. \]

This distinction becomes important when rational expressions are simplified.

Common Errors in Expansion

Failing to Multiply Every Term

Incorrect:

\[ 2(x+5)=2x+5. \]

Correct:

\[ 2(x+5)=2x+10. \]

Sign Errors

Incorrect:

\[ (x-3)(x+2)=x^2+5x-6. \]

Correct expansion:

\[ x^2+2x-3x-6=x^2-x-6. \]

Incorrect Squaring

Incorrect:

\[ (a+b)^2=a^2+b^2. \]

Correct:

\[ (a+b)^2=a^2+2ab+b^2. \]

Common Errors in Factoring

Ignoring the Greatest Common Factor

For

\[ 6x^2+12x, \]

writing only

\[ 2(3x^2+6x) \]

is a valid partial factorization, but

\[ 6x(x+2) \]

reveals the full common factor.

Using the Wrong Signs

If

\[ (x+a)(x+b)=x^2+(a+b)x+ab, \]

then both the sum and product of \(a\) and \(b\) must match the polynomial.

Stopping Too Early

For

\[ 3x^2-27, \]

factoring out \(3\) gives

\[ 3(x^2-9), \]

but the difference of squares can still be factored:

\[ 3(x-3)(x+3). \]

Worked Example: Expand and Simplify

Expand:

\[ (3x-2)(x+5). \]

Distribute:

\[ 3x(x+5)-2(x+5). \]

Then

\[ 3x^2+15x-2x-10. \]

Combine like terms:

\[ 3x^2+13x-10. \]

Therefore,

\[ (3x-2)(x+5)=3x^2+13x-10. \]

Worked Example: Factor a Trinomial

Factor:

\[ x^2-3x-10. \]

We need two numbers whose product is \(-10\) and whose sum is \(-3\).

The numbers are \(-5\) and \(2\).

Therefore,

\[ x^2-3x-10=(x-5)(x+2). \]

Check:

\[ (x-5)(x+2)=x^2-3x-10. \]

Worked Example: Factor Completely

Factor:

\[ 4x^3-36x. \]

First extract the greatest common factor:

\[ 4x(x^2-9). \]

Then factor the difference of squares:

\[ 4x(x-3)(x+3). \]

So the complete factorization over the integers is

\[ 4x(x-3)(x+3). \]

Worked Example: Choose the Better Form

Suppose we want to solve

\[ x^2+8x+15=0. \]

The expanded form does not immediately reveal the solutions.

Factor:

\[ (x+3)(x+5)=0. \]

Now the solutions are visible:

\[ x=-3 \]

or

\[ x=-5. \]

Here the factored form is more useful because the problem asks for zeros.

Practice

1. Expand

Expand and simplify:

\[ (x+4)(x+6). \]
Check Answer \[ x^2+6x+4x+24=x^2+10x+24. \]

2. Expand with Coefficients

Expand:

\[ (2x-1)(x+3). \]
Check Answer \[ 2x^2+6x-x-3=2x^2+5x-3. \]

3. Factor the Greatest Common Factor

Factor:

\[ 15x^2+20x. \]
Check Answer \[ 15x^2+20x=5x(3x+4). \]

4. Factor a Trinomial

Factor:

\[ x^2+9x+20. \]
Check Answer \[ x^2+9x+20=(x+4)(x+5). \]

5. Difference of Squares

Factor:

\[ x^2-64. \]
Check Answer \[ x^2-64=(x-8)(x+8). \]

6. Perfect Square Trinomial

Factor:

\[ x^2-12x+36. \]
Check Answer \[ x^2-12x+36=(x-6)^2. \]

7. Factor Completely

Factor:

\[ 2x^3-18x. \]
Check Answer

First factor out \(2x\):

\[ 2x(x^2-9). \]

Then factor the difference of squares:

\[ 2x(x-3)(x+3). \]

Key Ideas

Where This Leads

Expansion and factoring give us two complementary ways to move between additive and multiplicative structure.

Next, we will use these equivalence tools to study Rearranging Formulas, where equations containing several variables are transformed to isolate whichever quantity we need.