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Factoring Simple Quadratics

A free Algebra II lesson from the “Polynomial Factorization” unit, with a worked example and practice problems including step-by-step solutions.

Factoring quadratics rewrites a trinomial as a product of two binomials. This is useful because a product is easier to solve when it equals zero: if one factor is zero, the whole product is zero. For expressions like x^2 + bx + c, look for two numbers that multiply to c and add to b. When the leading coefficient is not 1, also track how the first terms multiply to the x^2 term. A common mistake is finding numbers that multiply correctly but do not add to the middle coefficient. Always multiply the factors back to check.

What you'll learn

Why it matters: Profit equations, projectile zero-crossings, and area-product problems factor into two binomials. Once an expression is written as (x − a)(x − b), the zero-product property hands you the solutions almost for free.

Worked example

Problem. Factor x^2 + 7x + 10.

  1. Find two numbers that multiply to 10.
  2. 5 and 2 multiply to 10 and add to 7.
  3. Write the factors as (x + 5)(x + 2).

Answer: (x + 5)(x + 2)

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Factor x^2 + 5x + 6.

Choices: (x + 2)(x + 3) · (x + 1)(x + 6) · (x - 2)(x - 3) · (x + 5)(x + 1)

Show solution
  1. 2 and 3 multiply to 6.
  2. 2 and 3 add to 5.

Answer: (x + 2)(x + 3)

2. For x^2 - 9x + 20, what two positive numbers multiply to 20 and add to 9? Enter the larger one.

Show solution
  1. Factor pairs of 20 include 1 and 20, 2 and 10, 4 and 5.
  2. 4 + 5 = 9.
  3. The larger number is 5.

Answer: 5

3. Factor x^2 - x - 12.

Choices: (x - 4)(x + 3) · (x + 4)(x - 3) · (x - 6)(x + 2) · (x - 12)(x + 1)

Show solution
  1. -4 and 3 multiply to -12 and add to -1.

Answer: (x - 4)(x + 3)

4. Solve x^2 - 6x + 8 = 0.

Choices: 2 and 4 · -2 and -4 · 1 and 8 · 3 and 5

Show solution
  1. Factor as (x - 2)(x - 4).

Answer: 2 and 4

5. Factor 2x^2 + 7x + 3.

Choices: (2x + 1)(x + 3) · (2x + 3)(x + 1) · (x + 1)(x + 6) · (2x - 1)(x - 3)

Show solution
  1. The middle terms are 6x and x.

Answer: (2x + 1)(x + 3)

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