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Factoring Polynomials with Patterns and Grouping: Algebra 1 Math Lesson

Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Algebra 1 Math Lessons – Mathbot, the Math Made Easy robot coach, with free printable lessons for Algebra 1

Looking for algebra 1 factoring polynomials? Then this Algebra 1 lesson on Factoring Polynomials with Patterns and Grouping explains each idea step by step, with worked examples.

Quick answer: Factoring rewrites a polynomial as a product of simpler polynomials. Start by taking out the greatest common factor, then look for a pattern: a difference of squares, a perfect square trinomial, a trinomial with two integers that fit, or grouping. Factoring helps solve equations because a product is zero only when one of its factors is zero.

Standards: A-SSE.A.2, A-SSE.B.3a · Level: Grade 9 Algebra 1

What does it mean to factor a polynomial?

Factoring rewrites a sum as a product. Because 6x² + 9x = 3x(2x + 3), the product form shows the pieces that multiply together, and it shows where a graph crosses the x-axis.

Greatest common factor (GCF)

The largest monomial that divides every term. Take it out first, every time: 12x³ + 18x² = 6x²(2x + 3).

1. Take out the greatest common factor2. Two terms? Try a difference of squares3. Three terms? Use a trinomial pattern4. Four terms? Try grouping5. Multiply to check the answer
A factoring checklist: always start with the GCF.

How do you factor x² + bx + c?

Find two integers whose product is c and whose sum is b. For x² + 7x + 12 the pair 3 and 4 works, so the factors are (x + 3)(x + 4).

x²4xx3x123x4
x² + 7x + 12 = (x + 3)(x + 4)

How do you factor when the leading coefficient is not 1?

Method: split the middle term

  1. Multiply a and c. For 6x² + 11x − 10 that is −60.
  2. Find two numbers with product −60 and sum 11: 15 and −4.
  3. Split: 6x² + 15x − 4x − 10.
  4. Group and factor each pair: 3x(2x + 5) − 2(2x + 5).
  5. Write the shared factor once: (2x + 5)(3x − 2).

Which patterns factor quickly?

Special patterns

a² − b² = (a + b)(a − b)

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

Four terms often factor by grouping: x³ + 4x² + 3x + 12 = x²(x + 4) + 3(x + 4) = (x² + 3)(x + 4).

Common mistake

A sum of squares such as x² + 16 does not factor over the real numbers, so do not write (x + 4)².

How does factoring find zeros?

If a product equals zero, one of its factors is zero (the zero-product property). To solve x² − 2x − 8 = 0, factor to (x − 4)(x + 2) = 0, so x = 4 or x = −2. These are the x-intercepts of the graph.

-4-3-2-1123456-10-8-6-4-22468xy(−2, 0)(4, 0)vertex (1, −9)
Zeros of y = (x − 4)(x + 2) are x = −2 and x = 4.
Mathbot’s tip

Always multiply your factors back out. If you do not get the original expression, one sign is wrong.

Key points to remember

  • Take out the greatest common factor before anything else.
  • For x² + bx + c, find two integers with product c and sum b.
  • a² − b² = (a + b)(a − b) and a² ± 2ab + b² = (a ± b)².
  • Four terms often factor by grouping in pairs.
  • Zero-product property: if (x − r)(x − s) = 0 then x = r or x = s.

Factoring Polynomials with Patterns and Grouping: frequently asked questions

What should you try first when factoring?

Look for a greatest common factor of all the terms and take it out. Doing this first makes the remaining polynomial smaller and easier to factor.

How do you factor x² + bx + c?

Find two integers that multiply to c and add to b. Then write (x + p)(x + q) using those two integers. Multiply back to check the middle term.

Why can you not factor x² + 25?

A sum of two squares has no pair of real numbers that multiply to 25 and add to 0. Only a difference of squares, such as x² − 25, factors.

What is the zero-product property?

If two numbers multiply to give zero, at least one of them is zero. For (x − 3)(x + 7) = 0, set each factor equal to zero to get x = 3 or x = −7.

How can you check a factoring answer?

Multiply the factors using FOIL or an area model. The product must match the original polynomial exactly, including every sign.