Factor Common Term: Reverse the Distributive Property

This guide sits inside the Algebra study map, where you can see the neighboring moves, models, and next-step guides that connect this principle to the rest of algebra.

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ

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The Principle

The move: Identify a factor present in every term of a sum, divide each term by it, and write the result as that factor multiplied by a parenthesized sum of the remainders.

The invariant: This produces an equivalent expression with the same value for every allowed variable assignment — factoring out a common term rewrites but does not evaluate or alter the expression.

Pattern: $ax + ay \quad \longrightarrow \quad a(x + y)$

Legal ✓	Illegal ✗
$6x + 9 \to 3(2x + 3)$	$6x + 9 \not\to 3(2x + 9)$

In the illegal case the factor $3$ was divided out of the first term but not out of the second. The second term inside the parentheses must also be divided by $3$: $9 \div 3 = 3$, giving $3(2x + 3)$.

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Conditions of Applicability

Condition: Terms share a common factor

Before applying, check: Inspect every term — the candidate factor must appear as a multiplicative factor in each one. For monomials, this means the coefficients divide appropriately and the required variable powers are present in each term. For non-monomial factors (e.g. a binomial like $(x+1)$), the entire expression must appear as a factor in every term. If even one term does not contain the factor, the move is not yet legal.

• $12x^2 + 8x$: valid — both terms share factor $4x$: $4x(3x + 2)$.
• $12x^2 + 8x + 5$: only $4x$ is shared by the first two terms, not by $5$. A factor of $4x$ cannot be pulled from all three terms; partial factoring or grouping must be used instead.
• $x^2 + 1$: no nontrivial common factor exists, so this move does not simplify the expression.

Want the complete framework behind this guide? Read Masterful Learning.

This move depends on the reverse view of Distributive Property. Compare it with Difference of Squares when a shared factor is not the real structure to exploit, and use it next with Zero Product when factoring turns an equation into separate factors.

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Common Failure Modes

Failure mode: dividing the factor out of only some of the terms and leaving the others unchanged → the expression inside the parentheses is wrong and the equality is broken.

Debug: after factoring, verify by redistributing: multiply the factor back through every term inside the parentheses and confirm you recover the original expression exactly.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What does “common factor” mean for a numeric coefficient versus a variable factor? How do you find the greatest common factor of two monomials?
• Why must the factor be divided out of every term — not just the first — for the rewrite to be valid?

For the Principle

• How do you decide how much to factor out: just the numeric GCF, the full monomial GCF, or something else? What guides that choice?
• What is your check that the factored form is correct before moving to the next step?

Between Principles

• How does Factor Common Term relate to the Distributive Property? When is each the appropriate direction to move?

Generate an Example

• Write a three-term polynomial where the numeric GCF and a variable factor are both present, and factor it completely. Identify what would go wrong if you factored out only the numeric GCF and stopped.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Identify a factor shared by every term, divide each term by it, and write the result as that factor times a parenthesized sum of the remainders.

Write the canonical pattern: _____$ax + ay = a(x + y)$

State the canonical condition: _____Terms share a common factor

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $12x^3 - 8x^2 + 4x$, factor out the greatest common monomial factor.

Step	Expression	Operation
0	$12x^3 - 8x^2 + 4x$	—
1	$12x^3 - 8x^2 + 4x$	Common monomial GCF identified as $4x$: $\gcd(12,8,4) = 4$; minimum variable power is $x^1$
2	$4x(3x^2 - 2x + 1)$	Divide each term by $4x$: $12x^3/4x = 3x^2$, $-8x^2/4x = -2x$, $4x/4x = 1$

Check (distribute back): $4x \cdot 3x^2 = 12x^3$, $4x \cdot (-2x) = -8x^2$, $4x \cdot 1 = 4x$ ✓

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Drills

Format A — Forward step

Apply Factor Common Term once. Pull out the greatest common monomial factor.

$6x + 9$

Reveal

GCF of 6 and 9 is 3; no variable is shared:

$3(2x + 3)$

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Apply Factor Common Term once.

$10y^2 + 15y$

Reveal

GCF: $5y$ (numeric GCF 5; lowest variable power $y^1$):

$5y(2y + 3)$

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Apply Factor Common Term once.

$-4x^3 + 12x^2$

Reveal

GCF of $|-4|$ and $|12|$ is $4$; lowest power of $x$ is $x^2$. Factor out $4x^2$ (keep leading sign positive inside):

$4x^2(-x + 3)$

Equivalently: $-4x^2(x - 3)$. Either form is valid; the latter is conventional when all terms have the same sign pattern.

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Apply Factor Common Term once.

$a^2b + ab^2$

Reveal

GCF: $ab$ (both $a$ and $b$ appear to at least the first power in each term):

$ab(a + b)$

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Format E — Canonicalization

Rewrite in fully factored form (factor out the greatest common monomial).

$18x^4 - 12x^3 + 6x^2$

Reveal

GCF: $\gcd(18,12,6) = 6$; lowest power of $x$ is $x^2$. GCF = $6x^2$:

$6x^2(3x^2 - 2x + 1)$

Check: $6x^2 \cdot 3x^2 = 18x^4$, $6x^2 \cdot 2x = 12x^3$, $6x^2 \cdot 1 = 6x^2$ ✓

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Rewrite in fully factored form.

$p^3q^2 + p^2q^3 - p^2q^2$

Reveal

GCF: $p^2q^2$ (lowest powers: $p^2$, $q^2$):

$p^2q^2(p + q - 1)$

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Identify which terms are eligible and factor only those. What do you do with the rest?

$9x^2 + 6x + 4$

Reveal

Check for a common factor across all three terms: $\gcd(9,6,4)$. $4$ is not divisible by $3$, so there is no common numeric factor greater than $1$. No variable appears in the constant term.

This expression cannot be factored by pulling out a common monomial. The move is not legal here. (It may be factorable as a trinomial by inspection or using the quadratic formula, but that is a different principle.)

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Reject the invalid rewrite. What is wrong?

$2x^2 + 6x \not\to 2(x^2 + 6x)$

Reveal

The factor $2$ divides $2x^2$ to give $x^2$ and divides $6x$ to give $3x$ — not $6x$. The corrected factored form is:

$2(x^2 + 3x)$

The error was dividing the first term by $2$ to get $x^2$ but then copying $6x$ directly rather than also dividing $6x \div 2 = 3x$. Always verify by redistributing every term inside the parentheses.

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Near-miss: the symbols look similar. Is this a legal Factor Common Term rewrite?

$xy + xz = x(y + z) \quad \text{vs.} \quad xy + wz \not\to x(y + z)$

Reveal

• $xy + xz = x(y + z)$: legal — $x$ is a factor of both $xy$ and $xz$.
• $xy + wz \not\to x(y + z)$: illegal — $w \neq x$, so $x$ is not a factor of $wz$. The two expressions look superficially similar (two two-letter products), but the second term must contain the candidate factor for the move to apply.

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Apply Factor Common Term to set up for the Zero Product Property.

$x^2 - 5x = 0$

Reveal

Both terms on the left share the factor $x$:

$x(x - 5) = 0$

This prepares the equation for the Zero Product Property.

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Apply Factor Common Term once.

$14a^2b - 21ab^2$

Reveal

GCF: $7ab$ (numeric GCF 7; both $a$ and $b$ appear to first power or higher in each term):

$7ab(2a - 3b)$

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Apply Factor Common Term once.

$(x + 1)^2 + 3(x + 1)$

Reveal

The expression $(x + 1)$ appears as a factor in both terms:

$(x + 1)(x + 1 + 3) = (x + 1)(x + 4)$

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Reject the invalid rewrite. What is wrong?

$8x^2 + 12x + 6 \not\to 2x(4x + 6 + 3)$

Reveal

The constant term $6$ does not contain the factor $x$, so $2x$ cannot be pulled from all three terms — the condition fails for the last term.

A valid greatest common monomial factor is $2$ only:

$2(4x^2 + 6x + 3)$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $3x^3 + 6x^2 - 9x$, rewrite the expression by factoring out the greatest common monomial factor.

Full solution

Step	Expression	Move
0	$3x^3 + 6x^2 - 9x$	—
1	$3x(x^2 + 2x - 3)$	Factor out GCF $3x$: $3x^3/3x = x^2$, $6x^2/3x = 2x$, $-9x/3x = -3$

Verify by distributing: $3x \cdot x^2 = 3x^3$, $3x \cdot 2x = 6x^2$, $3x \cdot (-3) = -9x$ ✓

The trinomial $x^2 + 2x - 3$ contains no further common monomial factor, so the target form is reached. To find the zeros of the original expression, the next step would be trinomial factoring followed by the Zero Product Property.

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FAQ

What is Factor Common Term?

Factor Common Term is the algebraic move of pulling a shared factor out of every term in a sum, rewriting the sum as that factor multiplied by a parenthesized expression. It is the reverse of the Distributive Property and preserves the value of the expression for every variable assignment.

When is Factor Common Term valid?

The move is valid when every term in the sum contains the candidate factor — meaning the factor appears as a multiplicative factor in every term. For monomial factors, the coefficients must divide appropriately and the required variable powers must be present in each term. For non-monomial factors, the entire expression must appear as a factor in each term. If even one term does not contain the factor, the move cannot be applied globally (though partial factoring by grouping may still be possible).

What goes wrong if I apply it to terms that don’t all share the factor?

You break the equality: the reorganized expression no longer equals the original. The fastest check is to redistribute after factoring — multiply the factor back through every interior term and verify you recover the original.

How is Factor Common Term different from the Distributive Property?

They are inverses: the Distributive Property expands $a(x + y)$ into $ax + ay$; Factor Common Term compresses $ax + ay$ into $a(x + y)$. Both preserve value. The distributive direction expands; the factoring direction reveals structure for further steps.

Does Factor Common Term work when the common factor is a variable expression?

Yes. The factor can be any expression — a number, a variable monomial, or a more complex expression — as long as it divides every term without remainder. For example, $(x+1)$ is a valid factor to pull from $(x+1)^2 + 3(x+1)$, giving $(x+1)(x+1+3) = (x+1)(x+4)$.

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How This Fits in Unisium

Factor Common Term sits at the gateway to factoring in the algebra progression: it follows the Distributive Property (its logical inverse) and is a prerequisite for the Zero Product Property. Building fluency with common-factor recognition — from numeric and monomial GCFs to larger expression factors — eliminates the most common bottleneck in polynomial solving. In Unisium, this move is practiced through short state-transition drills that train condition recognition: students learn to ask “does every term share this factor?” before applying, and to verify by redistributing afterward.

Explore further:

• Distributive Property — The expansion direction that Factor Common Term reverses
• Zero Product Property — The downstream principle Factor Common Term unlocks
• Elaborative Encoding — Build deep understanding of why the common-factor condition is necessary
• Retrieval Practice — Make the pattern and condition instantly accessible

Ready to master Factor Common Term? Start practicing with Unisium or explore the full learning framework in Masterful Learning.

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