1
Factoring is multiplication in reverse
Every factorisation can be verified by expanding it. This is the only topic on the path where checking your answer is faster than producing it, so there is never a reason to submit an unchecked one.
Lesson · Polynomials and Factoring · Factoring
Factoring Out the Greatest Common Factor
Factoring is the distributive law read right to left. The greatest common factor is what every term shares — the gcd of the coefficients and the lowest power of each common variable — and taking it out first is what keeps every later technique usable.
ab + ac = a(b + c) the distributive law, backwards
GCF = gcd of the coefficients × lowest power of each common variable
12x⁴y − 18x³y² + 6x³y = 6x³y(2x − 3y + 1)
One law, one recipe
Core ideas
What this lesson establishes
Factoring asks a different kind of question from anything so far: not "what does this equal" but "what would multiply to give this". The answer is a product, and it is checkable in seconds.
2
The GCF must divide EVERY term
Not most of them. A factor shared by two terms out of three is not a common factor of the polynomial, and pulling it out anyway leaves a fraction behind.
3
Take it out first, always
Every grouping, trinomial and special-form technique assumes the easy factor is gone. Skipping this step is what turns a one-line problem into a trinomial with coefficients too large to search sensibly.
The material
Finding what every term shares
What factoring means, how the GCF is computed, and the two details that go wrong.
Definition · Factoring
To factor a polynomial is to write it as a product of two or more polynomials. A factorisation is complete when no factor can be factored further.
x² + 5x factors as x(x + 5). The result is not simpler to look at — it is more useful, because a product tells you when the whole thing is zero, and a sum does not.
The licence is the distributive law, ab + ac = a(b + c), used from right to left. Nothing else is happening, and nothing else is needed.
Definition · Greatest common factor
For terms with integer coefficients, the GCF is the largest monomial that divides all of them. Its coefficient is the greatest common divisor of the coefficients, and each variable appears to the lowest power with which it appears in every term.
A variable that is missing from even one term is missing from the GCF entirely.
The integer-coefficient convention matters. Over the rational numbers any nonzero rational can be moved in or out of a product, so there is no useful largest numerical factor without first normalising the coefficients. Problems in this course use integer coefficients for the GCF decision; exact fractional coefficients can first be cleared by a common denominator.
Take 12x⁴y − 18x³y² + 6x³y. The coefficients are 12, 18, 6, whose gcd is 6. Every term has an x, to powers 4, 3, 3, so x³ is shared. Every term has a y, to powers 1, 2, 1, so y is shared. The GCF is 6x³y.
The word lowest is the one to hold on to. x³ is common because a term with only three xs cannot supply four; the shared part is limited by the poorest term, exactly as it is with the coefficients.
The term that leaves a 1 behind
When a term is the GCF, dividing it by the GCF leaves 1, and that 1 must be written:
5x² + 5x = 5x(x + 1) not 5x(x)
12x⁴y − 18x³y² + 6x³y = 6x³y(2x − 3y + 1)
Dropping the 1 changes the polynomial. Expand 5x(x) and you get 5x², which is not what you started with — and the expansion check catches it immediately.
Factoring out a negative
When the leading coefficient is negative it is usually worth taking the minus sign out with the GCF, because the grouping, trinomial and special-form techniques to come are stated for a positive leading term:
−4x³ + 8x² − 12x = −4x(x² − 2x + 3)
Every sign inside the bracket flips, for the same reason as in “Adding and Subtracting Polynomials”. Check by expanding: −4x · x² = −4x³, −4x · (−2x) = +8x², −4x · 3 = −12x. Correct.
Theorem · Why "greatest" matters
If you factor out a common factor that is not the greatest, the remaining bracket still contains a common factor.
12x² + 18x = 2x(6x + 9) is a true statement and an incomplete factorisation: 6x + 9 still has a 3 in it. The full answer is 6x(2x + 3).
Example · A factor that is not common
In 6x³ + 9x² + 4x the coefficients 6, 9, 4 have gcd 1, and although two terms carry x² the third has only x. The GCF is x, giving x(6x² + 9x + 4) — and no more. Pulling out 3x would leave 2x² + 3x + 4/3; that equality is true, but 3 is not the integer GCF because it does not divide the coefficient 4.
Interactive
The common factor
Divisor lists, intersected
Method
Taking out the greatest common factor
Coefficients and variables are handled separately, then combined.
1
Find the gcd of the coefficients
Ignore the variables entirely and take the greatest common divisor of the numbers, signs aside. If it is 1, the GCF has no numerical part.
2
For each variable, take the lowest power that appears in every term
Every term, not most terms. A variable absent from one term does not appear in the GCF at all.
3
Divide each term by the GCF to get the bracket
Term by term. Where a term equals the GCF exactly, the quotient is 1 and you write it. The bracket must have as many terms as the original.
4
Expand to check, then keep going
Multiply back out; it must reproduce the original exactly. Then look at the bracket — the GCF is the first step of the decision list, not the last, and what remains is usually factorable further.
Common factors
Score 0 / 0Choose the answer you can justify.
A three-term GCF, in full
12x⁴y − 18x³y² + 6x³y. Coefficients, then x, then y.
coefficients gcd(12, 18, 6) = 6
x powers 4, 3, 3 → lowest is x³
y powers 1, 2, 1 → lowest is y
GCF 6x³y
divide each term by 6x³y:
12x⁴y / 6x³y = 2x
−18x³y² / 6x³y = −3y
6x³y / 6x³y = 1 <-- write the 1
answer 6x³y(2x − 3y + 1)
check 6x³y·2x = 12x⁴y
6x³y·(−3y) = −18x³y²
6x³y·1 = 6x³y all three match
Three terms in, three terms in the bracket. That count is the fastest check that nothing was dropped, and the third term — the 1 — is the one at risk.
Had the y been missing from any single term, the GCF would have been 6x³ and the bracket would have carried the ys instead. The poorest term always decides.
For a faded pass, start −18x⁴ + 30x³ − 12x² with the supplied choice to take the negative out so the bracket begins positive. Find the numerical gcd and lowest power of x, divide all three terms, then expand. The result is −6x²(3x² − 5x + 2); a bracket with only two terms has dropped the quotient 2 from the final term.
Common mistakes
1
Dropping the 1 when a term is the GCF5x² + 5x = 5x(x + 1). Writing 5x(x) silently deletes a term, and the expansion check finds it in one line.
2
Taking out a factor that is missing from one termA variable or number must divide every term, and the two cases fail differently. Pulling x² out of 6x³ + 9x² + 4x leaves 6x + 9 + 4x⁻¹, whose negative exponent makes the bracket not a polynomial. Pulling a 3 out is an algebraically true rational rescaling, but not an integer GCF: 3 does not divide the coefficient 4, and the resulting 4/3 shows that the integer-coefficient normalisation was left.
3
Stopping at a common factor that is not the greatest12x² + 18x = 2x(6x + 9) leaves a 3 inside. Always look at the bracket after the step and ask whether it still has something in common.
Check your understanding
✓
Finish when the GCF is automatic and the bracket is always checked.
Given any integer-coefficient polynomial you should take the GCF out without thinking about it, keep the term count right, and then look at what remains and ask which technique comes next. Grouping, trinomial methods and special forms all begin from the bracket this step produces.
Note: The lab puts this first in its decision list for a reason worth stating plainly: a trinomial like 6x² + 15x + 6 looks like hard work until the 3 comes out and leaves 2x² + 5x + 2. The same problem, with a search space several times smaller.