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How do you factor a quadratic trinomial, and how do the factoring patterns (GCF, difference of squares, trinomials) fit together?

Factor quadratic expressions, including GCF, difference of squares, and trinomials, to reveal zeros and equivalent forms (Ohio A-SSE.3a, A-APR).

An Ohio Algebra I answer on factoring (A-SSE.3a): the GCF first, the difference of squares, factoring monic and non-monic trinomials by the product-sum method, and checking by expanding.

Generated by Claude Opus 4.812 min answer

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Jump to a section
  1. What this topic is asking
  2. Always start with the GCF
  3. Difference of squares
  4. Factoring a monic trinomial
  5. Factoring a non-monic trinomial
  6. How Ohio examines this topic
  7. Why "completely" is the word that matters
  8. How factoring connects to the graph
  9. Try this

What this topic is asking

Ohio standard A-SSE.3a asks you to factor a quadratic to reveal its zeros. Factoring is the reverse of multiplying, and it is the gateway to solving quadratics by the zero-product property. The patterns, GCF, difference of squares, and trinomial factoring, stack: you check for them in order. None of these patterns is on the reference sheet, so they must be memorized. This is a core skill in the Expressions and Equations reporting category and appears on Part 1.

Always start with the GCF

Before any other pattern, factor out the greatest common factor. This often turns a messy expression into a clean special pattern.

For 3x2βˆ’123x^2 - 12: the GCF is 33, giving 3(x2βˆ’4)3(x^2 - 4), and x2βˆ’4x^2 - 4 is a difference of squares, so 3(xβˆ’2)(x+2)3(x - 2)(x + 2).

Difference of squares

If the expression is one perfect square minus another, factor with the fixed pattern.

For x2βˆ’25=(xβˆ’5)(x+5)x^2 - 25 = (x - 5)(x + 5). Recognize perfect squares: 25=5225 = 5^2, 49=7249 = 7^2, 100=102100 = 10^2, and 4x2=(2x)24x^2 = (2x)^2.

Factoring a monic trinomial

A monic trinomial has leading coefficient 11: x2+bx+cx^2 + bx + c. Find two numbers that multiply to cc and add to bb.

The signs follow a rule: if c>0c > 0, both numbers share the sign of bb; if c<0c < 0, the numbers have opposite signs.

Factoring a non-monic trinomial

When a≠1a \neq 1, use the acac-method: multiply a⋅ca \cdot c, find two numbers that multiply to acac and add to bb, split the middle term, and factor by grouping.

How Ohio examines this topic

  • Equation response. Type the complete factorization.
  • Multiple choice. Choose the complete factorization, with "stopped too early" and "dropped the GCF" distractors.
  • Drag and drop. Build the factored form from given binomials.

Every factorization can be checked by expanding, which is the reliable way to catch a sign error on a no-calculator part.

Why "completely" is the word that matters

The test almost always asks you to factor completely, and partial factorizations are a frequent trap. "Completely" means continuing until no remaining factor can be broken down further. The classic miss is stopping at 2(x2βˆ’9)2(x^2 - 9) when x2βˆ’9x^2 - 9 is still a difference of squares, or factoring a trinomial but leaving a shared numerical factor inside a binomial. A safe routine is: pull the GCF, apply a special pattern or trinomial method, then re-examine each new factor to see whether it factors again. Only when every factor is irreducible are you done. This discipline is exactly what separates a fully credited answer from a "close but not complete" one on an exact-match item.

How factoring connects to the graph

Factoring is worth the effort because the factors hand you the zeros of the parabola y=ax2+bx+cy = ax^2 + bx + c for free. If x2βˆ’7x+12=(xβˆ’3)(xβˆ’4)x^2 - 7x + 12 = (x - 3)(x - 4), then the parabola crosses the xx-axis at x=3x = 3 and x=4x = 4, since the output is zero exactly when one factor is zero. So a single factorization answers "what are the solutions?", "where are the xx-intercepts?", and "what are the zeros of the function?" all at once. That is why factoring threads from this expressions module into solving quadratics and into graphing.

Try this

Q1. Factor x2+9x+20x^2 + 9x + 20 completely. [1 point]

  • Cue. 4β‹…5=204 \cdot 5 = 20, 4+5=94 + 5 = 9, so (x+4)(x+5)(x + 4)(x + 5).

Q2. Factor 3x2βˆ’273x^2 - 27 completely. [2 points]

  • Cue. GCF 33: 3(x2βˆ’9)=3(xβˆ’3)(x+3)3(x^2 - 9) = 3(x - 3)(x + 3).

Exam-style practice questions

Practice questions written in the style of ODEW exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Ohio Algebra I EOC (style)2 marksEquation response. Factor x2+2xβˆ’15x^2 + 2x - 15 completely.
Show worked answer β†’

The factored form is (x+5)(xβˆ’3)(x + 5)(x - 3).

Find two numbers that multiply to βˆ’15-15 (the constant) and add to 22 (the middle coefficient): +5+5 and βˆ’3-3 work, since 5β‹…(βˆ’3)=βˆ’155 \cdot (-3) = -15 and 5+(βˆ’3)=25 + (-3) = 2. So x2+2xβˆ’15=(x+5)(xβˆ’3)x^2 + 2x - 15 = (x + 5)(x - 3). Check by expanding: x2βˆ’3x+5xβˆ’15=x2+2xβˆ’15x^2 - 3x + 5x - 15 = x^2 + 2x - 15. The factors (x+5)(x + 5) and (xβˆ’3)(x - 3) reveal the zeros x=βˆ’5x = -5 and x=3x = 3.

Ohio Algebra I EOC (style)1 marksMultiple choice. Which is the complete factorization of 2x2βˆ’182x^2 - 18? (A) 2(x2βˆ’9)2(x^2 - 9) (B) 2(xβˆ’3)(x+3)2(x - 3)(x + 3) (C) (2xβˆ’6)(x+3)(2x - 6)(x + 3) (D) (xβˆ’3)(x+3)(x - 3)(x + 3)
Show worked answer β†’

The correct answer is (B).

Factor out the GCF 22 first: 2x2βˆ’18=2(x2βˆ’9)2x^2 - 18 = 2(x^2 - 9). Then x2βˆ’9x^2 - 9 is a difference of squares: 2(xβˆ’3)(x+3)2(x - 3)(x + 3). Option (A) stops too early (not fully factored), and (D) drops the GCF of 22 entirely. Complete factorization means continuing until no factor can be factored further.

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