The Central Idea

Distribution says

\[ a(b+c)=ab+ac. \]

Factoring reads the same equality in reverse:

\[ \boxed{ab+ac=a(b+c)}. \]

The expression on the left is written as a sum. The expression on the right is written as a product.

Both forms represent the same quantity.

Factoring Is Reverse Distribution

Consider

\[ 6x+12. \]

Both terms contain the common factor \(6\):

\[ 6x+12 = 6(x)+6(2). \]

Factor out the \(6\):

\[ \boxed{6x+12=6(x+2)}. \]

Expanding \(6(x+2)\) returns the original expression, so the two forms are equivalent.

What Is a Factor?

A factor is a quantity being multiplied.

In

\[ 6(x+2), \]

the two factors are

\[ 6 \]

and

\[ x+2. \]

In

\[ 3x(x-5), \]

the factors are

\[ 3,\qquad x,\qquad x-5. \]

What Is a Common Factor?

A common factor is a factor shared by every term in an expression.

Consider

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

Both terms contain a factor of \(4\), and both contain a factor of \(x\):

\[ 8x^2=4x(2x) \]

and

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

Therefore,

\[ \boxed{8x^2+12x=4x(2x+3)}. \]

The Greatest Common Factor

The greatest common factor, or GCF, is the largest factor shared by every term.

For

\[ 18x^3+24x^2, \]

the greatest common numerical factor of \(18\) and \(24\) is

\[ 6. \]

The variable parts \(x^3\) and \(x^2\) both contain

\[ x^2. \]

Therefore, the GCF is

\[ \boxed{6x^2}. \]

Factoring Out the GCF

Factor

\[ 18x^3+24x^2. \]

Divide each term by the GCF \(6x^2\):

\[ \frac{18x^3}{6x^2}=3x \]

and

\[ \frac{24x^2}{6x^2}=4. \]

Therefore,

\[ \boxed{18x^3+24x^2=6x^2(3x+4)}. \]

Finding the Numerical GCF

To factor numerical coefficients, find the greatest positive integer that divides every coefficient.

For example, in

\[ 30x^2+45x+75, \]

the coefficients are

\[ 30,\qquad45,\qquad75. \]

Their greatest common factor is

\[ 15. \]

Therefore,

\[ \boxed{30x^2+45x+75=15(2x^2+3x+5)}. \]

Finding the Variable GCF

When the same variable appears in every term, use the smallest exponent appearing among those terms.

Consider

\[ x^7+x^4+x^3. \]

The powers are

\[ x^7,\qquad x^4,\qquad x^3. \]

The smallest exponent is \(3\), so the greatest common variable factor is

\[ x^3. \]

Thus,

\[ \boxed{x^7+x^4+x^3=x^3(x^4+x+1)}. \]

Why We Use the Smallest Exponent

Consider

\[ x^5 \]

and

\[ x^3. \]

Both are divisible by \(x^3\):

\[ x^5=x^3x^2 \]

and

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

But \(x^4\) is not a factor of \(x^3\).

Therefore, the common variable factor cannot contain a power greater than the smallest exponent present.

Combining Numerical and Variable Factors

Consider

\[ 24x^5-36x^3+12x^2. \]

The numerical GCF is

\[ 12. \]

The smallest power of \(x\) is

\[ x^2. \]

Therefore, the full GCF is

\[ 12x^2. \]

Factoring gives

\[ \boxed{ 24x^5-36x^3+12x^2 = 12x^2(2x^3-3x+1) }. \]

Every Term Must Be Divided by the GCF

When factoring, the expression inside the parentheses is found by dividing every term by the common factor.

For

\[ 15x^4+10x^2-5x, \]

the GCF is

\[ 5x. \]

Divide term by term:

\[ \frac{15x^4}{5x}=3x^3, \] \[ \frac{10x^2}{5x}=2x, \]

and

\[ \frac{-5x}{5x}=-1. \]

Therefore,

\[ \boxed{15x^4+10x^2-5x=5x(3x^3+2x-1)}. \]

Factoring a Negative GCF

Sometimes it is useful to factor out a negative common factor.

Consider

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

We could factor out \(3x\):

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

But factoring out \(-3x\) gives

\[ \boxed{-3x(2x-3)}. \]

The second form is often preferred because the leading term inside the parentheses is positive.

Factoring Out \(-1\)

The simplest negative common factor is \(-1\).

For example,

\[ -x^2+5x-7 \]

can be written

\[ \boxed{-(x^2-5x+7)}. \]

Factoring out \(-1\) changes the sign of every term inside the parentheses.

Common Factors with Several Variables

The same logic applies when several variables are present.

Consider

\[ 18x^4y^3+24x^3y^5. \]

The numerical GCF is

\[ 6. \]

For \(x\), the smaller exponent is \(3\).

For \(y\), the smaller exponent is \(3\).

Therefore, the GCF is

\[ 6x^3y^3. \]

Factoring gives

\[ \boxed{ 18x^4y^3+24x^3y^5 = 6x^3y^3(3x+4y^2) }. \]

Example: Three Multivariable Terms

Factor

\[ 20a^4b^3-30a^3b^5+10a^2b^2. \]

The numerical GCF is

\[ 10. \]

The smallest power of \(a\) is

\[ a^2, \]

and the smallest power of \(b\) is

\[ b^2. \]

Therefore, the GCF is

\[ 10a^2b^2. \]

Thus,

\[ \boxed{ 20a^4b^3-30a^3b^5+10a^2b^2 = 10a^2b^2(2a^2b-3ab^3+1) }. \]

Factoring a Polynomial Factor

A common factor does not have to be a monomial.

Consider

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

Both terms contain the factor

\[ x+2. \]

Factor it out:

\[ \boxed{ 3x(x+2)+5(x+2) = (x+2)(3x+5) }. \]

This idea becomes especially important in factoring by grouping.

A Shared Binomial Can Behave Like a Common Number

Compare

\[ 3a+5a \]

with

\[ 3(x+1)+5(x+1). \]

In the first expression, \(a\) is common:

\[ 3a+5a=a(3+5). \]

In the second, the entire binomial \(x+1\) is common:

\[ 3(x+1)+5(x+1) = (x+1)(3+5). \]

Therefore,

\[ \boxed{3(x+1)+5(x+1)=8(x+1)}. \]

Always Look for a GCF First

Before trying any more specialized factoring method, inspect all terms for a common factor.

For example,

\[ 6x^3+12x^2+18x \]

has the GCF

\[ 6x. \]

Factoring it out gives

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

Even when the remaining polynomial will later require another factoring technique, extracting the GCF first simplifies the problem.

What Is Factoring by Grouping?

Sometimes no nontrivial factor is shared by every term, but groups of terms contain useful common factors.

Consider

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

There is no single variable factor common to all four terms.

But the first two terms share \(a\), and the last two share \(b\).

Group them:

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

The Basic Grouping Pattern

Starting with

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

factor each group:

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

Now both terms contain the common binomial

\[ x+y. \]

Factor again:

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

Grouping Works in Two Stages

Factoring by grouping usually has two levels.

  1. Factor a GCF from each group.
  2. Factor the common polynomial factor that appears afterward.

The goal of the first stage is to make the remaining grouped expressions identical.

Example: Factoring by Grouping

Factor

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

Group the first two and last two terms:

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

Factor each group:

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

Now \(x+3\) is common:

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

Why the Grouping Worked

In

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

the first pair factors as

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

while the second pair factors as

\[ 2(x+3). \]

The same binomial \(x+3\) appears in both.

That shared binomial becomes the next common factor.

Grouping with a Negative Factor

Sometimes one group produces the opposite of the binomial we need.

Factor

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

Group:

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

Factor the first group:

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

Factor \(-3\) from the second group:

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

Now factor the common binomial:

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

Why Factoring Out a Negative Can Be Necessary

Suppose one group contains

\[ -3x+12. \]

Factoring out \(3\) gives

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

But

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

Factoring out \(-3\) directly gives

\[ -3(x-4), \]

which matches the factor from the other group.

Choosing the sign strategically is part of recognizing structure.

Example: Coefficients in Grouping

Factor

\[ 6x^3+9x^2+4x+6. \]

Group:

\[ (6x^3+9x^2)+(4x+6). \]

Factor each group:

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

The common binomial is

\[ 2x+3. \]

Therefore,

\[ \boxed{ 6x^3+9x^2+4x+6 = (2x+3)(3x^2+2) }. \]

Grouping Does Not Always Use Adjacent Terms

Sometimes the original order does not reveal a useful grouping.

Consider

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

Grouping adjacent terms gives

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

which does not immediately create matching factors.

Reorder the terms:

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

Then

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

and finally

\[ \boxed{(a+b)(x+y)}. \]

Because addition is commutative, terms may be reordered when doing so reveals useful structure.

Example: Reordering Before Grouping

Factor

\[ x^3+6x+2x^2+12. \]

Reorder into descending powers:

\[ x^3+2x^2+6x+12. \]

Group:

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

Factor:

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

Therefore,

\[ \boxed{(x+2)(x^2+6)}. \]

Grouping with Several Variables

Factoring by grouping works with multivariable polynomials as well.

Consider

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

We already found

\[ \boxed{(a+b)(x+y)}. \]

The variables do not need to play different formal roles. The method depends only on common multiplicative structure.

Example: Multivariable Grouping

Factor

\[ 3x^2y+6xy^2+5xz+10yz. \]

Group:

\[ (3x^2y+6xy^2)+(5xz+10yz). \]

Factor the first group:

\[ 3xy(x+2y). \]

Factor the second group:

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

Now factor the common binomial:

\[ \boxed{ 3x^2y+6xy^2+5xz+10yz = (x+2y)(3xy+5z) }. \]

Factoring Completely

A polynomial is factored completely relative to the methods currently available when no further common factor can be extracted from the remaining factors.

Consider

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

Writing

\[ 3x^2(4x+6) \]

is a valid factorization, but it is not complete because \(4x+6\) still has a common factor of \(2\).

The complete GCF factorization is

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

A GCF May Remain After Grouping

Sometimes grouping reveals a product whose factors can be simplified further.

Factor

\[ 2x^3+4x^2+6x+12. \]

One grouping gives

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

Then

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

But

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

Therefore,

\[ \boxed{2(x+2)(x^2+3)}. \]

Always inspect the result for additional common factors.

Factoring Changes Form, Not Value

The expressions

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

and

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

are equivalent.

Factoring does not alter the polynomial function. It changes the representation so that multiplicative structure becomes visible.

Why Factored Form Is Useful

Factored form can reveal information that expanded form hides.

It can make it easier to:

Later lessons will use factoring for all of these purposes.

Factoring and the Zero-Product Principle

Factored form becomes especially powerful when a polynomial equation is equal to zero.

For example, if

\[ (x+2)(x-5)=0, \]

then at least one factor must be zero.

This observation will later connect factoring directly to solving polynomial equations.

For now, the main goal is learning how to reveal the factors correctly.

Checking a Factorization

The fastest general check is to distribute the factors and verify that the original polynomial returns.

Suppose we claim

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

Expand the right side:

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

The original polynomial is recovered, so the factorization is correct.

Checking a Grouping Factorization

Suppose

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

Expand:

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

This gives

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

Reorder:

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

The original polynomial returns.

Evaluating as an Additional Check

A numerical substitution can provide a quick secondary check.

Suppose

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

was factored as

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

At \(x=2\), the expanded form gives

\[ 8+12+4+6=30. \]

The factored form gives

\[ (2+3)(4+2)=5\cdot6=30. \]

Matching values support the calculation, although expansion remains the direct algebraic verification.

Common Error: Factoring Only Some Terms

Suppose we factor

\[ 12x^2+18x+6. \]

The GCF of all three terms is \(6\).

The correct factorization is

\[ \boxed{6(2x^2+3x+1)}. \]

The factor must divide every term, including the constant term.

Common Error: Using Too Large a Variable Factor

Consider

\[ x^5+x^3. \]

The GCF is

\[ x^3, \]

not \(x^5\), because \(x^5\) does not divide \(x^3\).

Therefore,

\[ \boxed{x^5+x^3=x^3(x^2+1)}. \]

Common Error: Forgetting the \(1\)

Consider

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

Factoring out \(x^2\) gives

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

The second term becomes \(1\) because

\[ \frac{x^2}{x^2}=1. \]

Omitting the \(1\) would give

\[ x^2(x)=x^3, \]

which loses the original \(x^2\) term.

Common Error: Changing Signs Incorrectly

If we factor \(-3\) from

\[ -6x+9, \]

every term inside changes sign:

\[ \boxed{-6x+9=-3(2x-3)}. \]

Expanding verifies the signs:

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

Common Error: Grouping Without Creating a Common Factor

Simply placing parentheses around terms does not accomplish factoring.

The purpose of grouping is to produce a repeated factor.

For

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

the useful grouping is

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

because it leads to

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

The repeated factor \(x+3\) is what makes the method succeed.

Common Error: Stopping Too Early

Consider

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

Factoring out \(2x^2\) gives

\[ 2x^2(4x+6). \]

This is correct but incomplete because \(4x+6\) still has a factor of \(2\).

The greater common factor is \(4x^2\), so

\[ \boxed{8x^3+12x^2=4x^2(2x+3)}. \]

Common Error: Expanding Instead of Factoring

If the instruction says to factor

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

the goal is to rewrite the sum as a product:

\[ \boxed{6x(x+2)}. \]

Factoring and expansion move in opposite directions.

A Reliable Procedure for GCF Factoring

  1. Find the greatest common numerical factor.
  2. For each variable shared by every term, choose the smallest exponent.
  3. Combine these pieces to form the GCF.
  4. Divide every term by the GCF.
  5. Write the GCF outside parentheses.
  6. Place the quotients inside the parentheses.
  7. Inspect the result for any remaining common factor.
  8. Expand to verify when needed.

A Reliable Procedure for Factoring by Grouping

  1. Check for a GCF across the entire polynomial first.
  2. Reorder terms if doing so exposes a useful pattern.
  3. Divide the terms into groups.
  4. Factor the GCF from each group.
  5. Look for an identical polynomial factor in every group.
  6. If necessary, factor out a negative number to make the factors match.
  7. Factor out the repeated polynomial factor.
  8. Inspect the remaining factor for further common factors.
  9. Expand the product to verify the result.

Example: Full GCF Workflow

Factor completely:

\[ 42x^5y^3-63x^3y^4+21x^2y^2. \]

Numerical coefficients:

\[ \gcd(42,63,21)=21. \]

Smallest power of \(x\):

\[ x^2. \]

Smallest power of \(y\):

\[ y^2. \]

Therefore, the GCF is

\[ 21x^2y^2. \]

Divide each term:

\[ \frac{42x^5y^3}{21x^2y^2}=2x^3y, \] \[ \frac{-63x^3y^4}{21x^2y^2}=-3xy^2, \] \[ \frac{21x^2y^2}{21x^2y^2}=1. \]

Therefore,

\[ \boxed{ 42x^5y^3-63x^3y^4+21x^2y^2 = 21x^2y^2(2x^3y-3xy^2+1) }. \]

Example: Full Grouping Workflow

Factor

\[ 2x^3+6x^2+5x+15. \]

There is no nontrivial GCF across all four terms.

Group:

\[ (2x^3+6x^2)+(5x+15). \]

Factor each group:

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

Factor the common binomial:

\[ \boxed{(x+3)(2x^2+5)}. \]

Expanding confirms:

\[ x(2x^2+5)+3(2x^2+5) = 2x^3+5x+6x^2+15, \]

which reorders to the original polynomial.

Practice 1: Numerical GCF

Factor completely:

\[ 24x+36. \]

Practice 2: Numerical and Variable GCF

Factor completely:

\[ 20x^4-30x^3+10x^2. \]

Practice 3: Negative GCF

Factor so that the leading term inside the parentheses is positive:

\[ -15x^3+20x^2. \]

Practice 4: Several Variables

Factor completely:

\[ 18a^4b^3+27a^2b^5. \]

Practice 5: Shared Binomial

Factor:

\[ 7x(x-4)+3(x-4). \]

Practice 6: Grouping

Factor:

\[ x^3+5x^2+2x+10. \]

Practice 7: Grouping with a Negative Factor

Factor:

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

Practice 8: Reorder and Group

Factor:

\[ 3x^3+8x+6x^2+16. \]

Practice 9: Factor Completely

Factor completely:

\[ 6x^3+12x^2+18x+36. \]

Practice 10: Identify the Error

A student writes

\[ 12x^3+18x^2=3x^2(4x+6) \]

and says the polynomial has been factored completely.

Explain what remains to be done.

Practice Answers

Practice 1

The GCF of \(24\) and \(36\) is \(12\):

\[ \boxed{24x+36=12(2x+3)}. \]

Practice 2

The numerical GCF is \(10\), and the smallest power of \(x\) is \(x^2\):

\[ \boxed{ 20x^4-30x^3+10x^2 = 10x^2(2x^2-3x+1) }. \]

Practice 3

Factor out \(-5x^2\):

\[ \boxed{-15x^3+20x^2=-5x^2(3x-4)}. \]

Practice 4

The GCF is \(9a^2b^3\):

\[ \boxed{ 18a^4b^3+27a^2b^5 = 9a^2b^3(2a^2+3b^2) }. \]

Practice 5

The common factor is \(x-4\):

\[ \boxed{ 7x(x-4)+3(x-4) = (x-4)(7x+3) }. \]

Practice 6

\[ (x^3+5x^2)+(2x+10) \] \[ = x^2(x+5)+2(x+5) \] \[ = \boxed{(x+5)(x^2+2)}. \]

Practice 7

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

This lesson has reached the grouping factorization. The remaining factor \(x^2-4\) is a special form that will be handled systematically in the next factoring lesson.

Practice 8

Reorder:

\[ 3x^3+6x^2+8x+16. \]

Group:

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

Therefore,

\[ \boxed{(x+2)(3x^2+8)}. \]

Practice 9

First factor the overall GCF \(6\):

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

Then group:

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

Therefore,

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

Practice 10

The factorization

\[ 3x^2(4x+6) \]

is correct but incomplete because \(4x+6\) still has a common factor of \(2\).

Factor that \(2\):

\[ 3x^2\cdot2(2x+3). \]

Therefore,

\[ \boxed{12x^3+18x^2=6x^2(2x+3)}. \]

What to Remember

Factoring reverses distribution:

\[ \boxed{ab+ac=a(b+c)}. \]

For GCF factoring:

For factoring by grouping:

The most important habit is simple: always check for a common factor before using a more specialized factoring method.

Where We Go Next

Common-factor methods reveal factors that are already visibly shared across terms. Many important polynomials require a different kind of pattern recognition.

The next lesson, Factoring Trinomials and Special Forms, develops methods for factoring quadratic trinomials, perfect-square trinomials, differences of squares, and other recurring polynomial structures.