factor.

Factoring Trinomials Calculator (Quadratic)

All steps are free — no sign-up, no paywall.

Integer coefficients · degree ≤4 · up to 2 of x/y/z/w/p/c/d · + − * ^ ( )

Examples & math keys
x^2+5x+62x^2+5x+3x^2-445w^4+30w^3+5w^2x^2+x+1

Use ^ for powers; 5x works too. Limits: 220 characters and coefficient magnitude up to 1 trillion.

Worked example

FACTORED FORM
(x+3)*(x+2)

from x^2+5*x+6

✓ Checked

Show steps

Trinomial x² + bx + c

  1. Find two numbers with the right sum and product

    m+n=5, m*n=6

    Let m and n be the missing numbers in (x+m)*(x+n). Expanding gives middle coefficient m+n and last coefficient m*n.

  2. Calculate the two numbers

    (5)^2-4*(6)=1^2 m=(5+1)/2=3 n=(5-1)/2=2 3+(2)=5; 3*(2)=6

    Their squared difference is the sum squared minus four times the product. Half the sum plus or minus that difference gives the two numbers directly.

  3. Write the two factors

    x^2+5*x+6 = (x+3)*(x+2)

    Substitute 3 and 2 for the two missing numbers. Their sum and product reproduce the original coefficients.

  4. Recognize a prime linear polynomial

    x+3

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  5. Recognize a prime linear polynomial

    x+2

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  6. Put the factors together

    (x+3)*(x+2)

    Combine the factors obtained in the preceding steps. Repeated factors are written as powers.

  7. Check by expanding

    (x+3)*(x+2) = x^2+2*x+3*x+6 = x^2+5*x+6

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓

How to Factor a Trinomial

First combine like terms and take out the greatest common factor. For a quadratic with leading coefficient 1, find two numbers whose sum is the middle coefficient and whose product is the constant. This calculator determines those numbers from their sum and difference, then checks both conditions.

For ax² + bx + c with a different leading coefficient, the AC method splits the middle term into two pieces whose coefficients multiply to ac and add to b. Factor each group and extract the shared binomial. Continue through repeated factors or a difference of squares, then expand to verify.

Common mistakes

  • Skipping the GCF: Take out the common number and variable powers first. Otherwise you may miss a repeated factor or stop before the expression is fully factored.
  • Checking only the product: The two numbers must have the required sum too. Check both signs against the middle coefficient.
  • Forgetting the leading coefficient: When a is not 1, use the product ac for the middle-term split, rather than c alone.
  • Forcing an answer: Some quadratics are prime over the integers. Keep that conclusion when the exact check rules out integer factors; always multiply any proposed factors back.

FAQ

How do you factor a trinomial?

Take out the GCF, then identify the leading coefficient. Use sum and product for x² + bx + c, or the AC method for ax² + bx + c; expand the resulting factors to check them.

Is a quadratic the same as a trinomial?

Quadratic describes degree two, while trinomial means three terms. Many expressions are both; this tool also handles a quadratic binomial such as x² − 4 with the difference-of-squares rule.

What if the trinomial cannot be factored?

A prime polynomial has no further factorization over the integers. The calculator states this only after an exact check; it does not invent factors or treat a failed calculation as proof.

Worked examples with steps

x^2+5x+6

FACTORED FORM
(x+3)*(x+2)

from x^2+5*x+6

✓ Checked

Show steps

Trinomial x² + bx + c

  1. Find two numbers with the right sum and product

    m+n=5, m*n=6

    Let m and n be the missing numbers in (x+m)*(x+n). Expanding gives middle coefficient m+n and last coefficient m*n.

  2. Calculate the two numbers

    (5)^2-4*(6)=1^2 m=(5+1)/2=3 n=(5-1)/2=2 3+(2)=5; 3*(2)=6

    Their squared difference is the sum squared minus four times the product. Half the sum plus or minus that difference gives the two numbers directly.

  3. Write the two factors

    x^2+5*x+6 = (x+3)*(x+2)

    Substitute 3 and 2 for the two missing numbers. Their sum and product reproduce the original coefficients.

  4. Recognize a prime linear polynomial

    x+3

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  5. Recognize a prime linear polynomial

    x+2

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  6. Put the factors together

    (x+3)*(x+2)

    Combine the factors obtained in the preceding steps. Repeated factors are written as powers.

  7. Check by expanding

    (x+3)*(x+2) = x^2+2*x+3*x+6 = x^2+5*x+6

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓

2x^2+5x+3

FACTORED FORM
(2*x+3)*(x+1)

from 2*x^2+5*x+3

✓ Checked

Show steps

AC method

  1. Multiply the first and last coefficients

    2*(3)=6; m+n=5

    Call the two missing numbers m and n. They must add to 5 and multiply to 6; this makes both groups share a factor after we split the middle term.

  2. Calculate the two numbers

    (5)^2-4*(6)=1^2 m=(5+1)/2=3 n=(5-1)/2=2 3+(2)=5; 3*(2)=6

    Their squared difference is the sum squared minus four times the product. Half the sum plus or minus that difference gives the two numbers directly.

  3. Split the middle term

    2*x^2+5*x+3 = 2*x^2+(3*x)+(2*x)+(3)

    3+(2)=5 and 3*(2)=6; splitting changes the grouping, not the value.

  4. Factor each group

    (2*x^2+3*x)+(2*x+3) = x*(2*x+3)+(1)*(2*x+3)

    Take the greatest common factor out of each pair; both groups now contain the same expression.

  5. Write the two factors

    2*x^2+5*x+3 = (2*x+3)*(x+1)

    Use the distributive property in reverse to take out the shared factor.

  6. Recognize a prime linear polynomial

    2*x+3

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  7. Recognize a prime linear polynomial

    x+1

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  8. Put the factors together

    (2*x+3)*(x+1)

    Combine the factors obtained in the preceding steps. Repeated factors are written as powers.

  9. Check by expanding

    (2*x+3)*(x+1) = 2*x^2+2*x+3*x+3 = 2*x^2+5*x+3

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓

x^2-4

FACTORED FORM
(x+2)*(x-2)

from x^2-4

✓ Checked

Show steps

Difference of squares

  1. Recognize a difference of squares

    x^2-4 = (x)^2-(2)^2

    Both terms are perfect squares, and they are subtracted: A^2-B^2=(A+B)(A-B).

  2. Multiply the conjugate factors

    x^2-4 = (x+2)*(x-2)

    The cross terms cancel when a sum and a difference are multiplied.

  3. Recognize a prime linear polynomial

    x+2

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  4. Recognize a prime linear polynomial

    x-2

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  5. Put the factors together

    (x+2)*(x-2)

    Combine the factors obtained in the preceding steps. Repeated factors are written as powers.

  6. Check by expanding

    (x+2)*(x-2) = x^2-2*x+2*x-4 = x^2-4

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓

45w^4+30w^3+5w^2

FACTORED FORM
5*w^2*(3*w+1)^2

from 45*w^4+30*w^3+5*w^2

✓ Checked

Show steps

GCF + Perfect square

  1. Take out the greatest common factor

    45*w^4+30*w^3+5*w^2 = 5*w^2*(9*w^2+6*w+1)

    Every term is divisible by this factor. Dividing each term by it leaves the expression in parentheses.

  2. Recognize a perfect-square trinomial

    9*w^2+6*w+1 = (3*w)^2+2*(3*w)*(1)+(1)^2

    The outside terms are squares, and the middle term is twice the product of their square roots.

  3. Write the repeated factor

    9*w^2+6*w+1 = (3*w+1)^2

    A^2+2AB+B^2=(A+B)^2 and A^2-2AB+B^2=(A-B)^2.

  4. Recognize a prime linear polynomial

    3*w+1

    A linear polynomial with no integer common factor cannot split into lower-degree polynomial factors.

  5. Put the factors together

    5*w^2*(3*w+1)^2

    Combine the factors obtained in the preceding steps. Repeated factors are written as powers.

  6. Check by expanding

    5*w^2*(3*w+1)^2 = 45*w^4+15*w^3+15*w^3+5*w^2 = 45*w^4+30*w^3+5*w^2

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓

x^2+x+1

PRIME POLYNOMIAL
(x^2+x+1)

from x^2+x+1

This polynomial cannot be factored over the integers (it is prime).

✓ Checked

Show steps

Irreducible over integers

  1. Check whether rational factors exist

    D=(1)^2-4*(1)*(1)=-3

    The discriminant is negative, so this quadratic has no real linear factors and therefore no rational linear factors.

  2. Check by expanding

    (x^2+x+1) = x^2+x+1

    Expanding and collecting like terms reproduces the original polynomial, so the factorization is equal to the input. ✓