Enter one polynomial, identify the greatest common factor first, and continue until every factor is irreducible over the stated number system.
Result
The greatest common numerical factor is 3 and both terms contain x, so the greatest common factor is 3x.
3x(2x−5)
Conditions
The expression is treated as a polynomial in x.
Coefficients are factored over the integers unless another domain is stated.
The final product is checked by expansion.
Steps
Find the numerical common factor The greatest common divisor of 6 and 15 is 3.gcd(6,15)=3
Find the variable common factor Both terms contain at least one factor of x.min(2,1)=1
Divide each term by the GCF Factoring out 3x leaves 2x from the first term and negative 5 from the second.6x2−15x=3x(2x−5)
Check for further factoring The linear factor 2x-5 is irreducible over the integers.3x(2x−5)
✓
Independent check
Expanding gives 3x(2x)-3x(5)=6x^2-15x.
Scope
What this factor covers
Factoring rewrites a polynomial as a product with the same value. Start with the greatest common factor, then test the remaining polynomial for a trinomial or special-product pattern, and expand the result to verify it.
Greatest common factors
Extract the largest coefficient and variable power shared by every term.
Examples: 6x^2-15x, 8a^3b+12a^2b^2
Quadratic trinomials
Find binomial factors whose product and middle terms reproduce the quadratic.
Examples: x^2+7x+12, 2x^2+7x+3
Difference of squares
Use a^2-b^2=(a-b)(a+b) after removing any common factor.
Examples: x^2-25, 9y^2-16
Perfect-square trinomials
Recognize matching square terms and twice their product in the middle.
Examples: x^2-6x+9, 4x^2+4x+1
How to use it
Enter enough information for one clear task
1
Write the polynomial in descending order
A consistent term order makes the degree, leading coefficient, and missing powers easier to see.
2
Take out the GCF
Check coefficients and variable powers across every term before looking for a special pattern.
3
Test the remaining structure
Look for a trinomial pair, difference of squares, or perfect square, then repeat until no factor can be reduced in the chosen domain.
4
Expand the product
Multiply the factors back together and compare every coefficient with the original polynomial.
Worked inputs
Examples to try
Use these examples to recognize the method, compare equivalent forms, and check your own work.
Monic trinomial
Find two integers with product 12 and sum 7.
x2+7x+12
Expected result
(x+3)(x+4)
Difference of squares
Recognize x squared minus 5 squared.
x2−25
Expected result
(x−5)(x+5)
Leading coefficient greater than one
Use factors that produce the middle terms 6x and x.
2x2+7x+3
Expected result
(2x+1)(x+3)
GCF followed by a special product
Factor 3x first, then factor the difference of squares.
3x3−12x
Expected result
3x(x−2)(x+2)
Perfect square with a negative middle term
Recognize x squared, 3 squared, and negative 2 times x times 3.
x2−6x+9
Expected result
(x−3)2
Perfect square with a nonunit coefficient
Recognize the square of 2x+1.
4x2+4x+1
Expected result
(2x+1)2
Complete example
Take out the GCF before factoring the trinomial
Every coefficient is divisible by 2. Removing that common factor makes the remaining monic trinomial easier to factor.
2x2−8x−10
1
Extract the greatest common factor
Divide each coefficient by 2.
2x2−8x−10=2(x2−4x−5)
2
Find the trinomial pair
Negative 5 and positive 1 multiply to negative 5 and add to negative 4.
(−5)(1)=−5,−5+1=−4
3
Write the complete factorization
Keep the GCF outside the two binomial factors.
2(x−5)(x+1)
2(x−5)(x+1)
Verification: First expand (x-5)(x+1)=x^2-4x-5, then multiply by 2 to recover 2x^2-8x-10.
Avoidable errors
Common mistakes and how to fix them
Skipping the greatest common factor
Problem: Stop after finding factors of the inner-looking terms.
Why it matters: The answer is not complete if every term still shares a factor.
Better approach: Check the numerical coefficients and the smallest variable exponent before any other method.
Choosing a pair with the wrong sum
Problem: Factor x^2+7x+12 as (x+2)(x+6).
Why it matters: The constants multiply to 12, but their sum is 8 rather than 7.
Better approach: Check both the product and the middle-term sum before accepting the pair.
Stopping before the product is fully factored
Problem: Leave 3x(x^2-4) as the final answer.
Why it matters: The remaining quadratic is still a difference of squares.
Better approach: Continue to 3x(x-2)(x+2) over the integers.
Confusing factors with solutions
Problem: Report x-3 and x+2 as roots without an equation equal to zero.
Why it matters: Factoring rewrites an expression; roots are values obtained only when a product is set equal to zero.
Better approach: Keep the factored expression unless the original task is to solve an equation.
Trust the result for the right reasons
Checks, assumptions, and limits
How results are checked
Every proposed factorization is expanded to compare all coefficients.
The greatest common factor is checked before a secondary pattern.
The stated coefficient domain determines when a factor is irreducible.
When to stop and revise the input
A polynomial irreducible over the integers may factor over the real or complex numbers.
The intended variable must be clear in an expression with several symbols.
Non-polynomial expressions require a different simplification or identity method.
Common questions
Factor a polynomial completely FAQ
What should I check before factoring a trinomial?
Look for a greatest common factor across every term. Removing it first makes the remaining coefficients smaller and prevents an incomplete final answer.
How do I know a factorization is correct?
Expand the proposed factors and combine like terms. The result must reproduce every term and coefficient of the original polynomial.
What does irreducible mean?
It means the polynomial cannot be written as a product of lower-degree nonconstant polynomials using coefficients from the stated number system. The answer can change when the domain changes.
Does factoring also solve the equation?
Only when the factored polynomial is set equal to zero. Then the zero-product property turns each factor into a candidate equation.
Why is x squared plus 1 not factored over the reals?
It has no real zeros, so it has no real linear factors. Over the complex numbers it factors as (x-i)(x+i).
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