How to Solve a Quadratic Equation by Factoring?

Factoring a quadratic equation is the fastest method for solving \(\color{blue}{\text{ ax }^{2} + \text{ bx } + c = 0}\) when it factors cleanly. The strategy is to rewrite the quadratic as a product of two binomials and then apply the Zero Product Property to find the solutions. This guide covers the most common factoring patterns you will encounter in Algebra 1.

What Is Factoring a Quadratic?

Factoring a quadratic means expressing \(\color{blue}{\text{ ax }^{2} + \text{ bx } + c}\) as \(\color{blue}{(\text{ px } + q)(\text{ rx } + s)}\). Once factored and set equal to zero, each factor can be solved independently: if \(\color{blue}{\text{ AB } = 0}\), then \(\color{blue}{A = 0}\) or \(\color{blue}{B = 0}\) (the Zero Product Property).

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Factoring Methods

Case 1: Leading \(\color{blue}{\text{ Coefficient } = 1}\) (x² + \(\color{blue}{\text{ bx } + c}\))

Find two numbers that multiply to \(\color{blue}{c}\) and add to \(\color{blue}{b}\). Those numbers are the constants in the binomial factors.

• \(\color{blue}{x^{2} + 5x + 6}\): need two numbers × = 6 and + = 5  →  2 and 3.
  Factored: \(\color{blue}{(x + 2)(x + 3)}\). Solutions: \(\color{blue}{x = -2, x = -3}\).

Case 2: Difference of Two Squares (\(\color{blue}{a^{2} – b^{2}}\))

Use the pattern: \(\color{blue}{a^{2} – b^{2} = (a + b)(a – b)}\).

• \(\color{blue}{x^{2} – 16 = (x + 4)(x – 4)}\). Solutions: \(\color{blue}{x = \pm 4}\).

Case 3: Leading Coefficient ≠ 1 (ax² + \(\color{blue}{\text{ bx } + c}\))

Find two numbers that multiply to \(\color{blue}{a \times c}\) and add to \(\color{blue}{b}\), then factor by grouping.

• \(\color{blue}{2x^{2} + 7x + 3}\): \(\color{blue}{a \times c = 6}\); need two numbers × = 6 and + = 7  →  1 and 6.
  Rewrite: \(\color{blue}{2x^{2} + x + 6x + 3}\)  →  \(\color{blue}{x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)}\).

Case 4: Perfect Square Trinomial

Recognize when the quadratic is \(\color{blue}{(x + k)^{2} = x^{2} + 2\text{ kx } + k^{2}}\).

• \(\color{blue}{x^{2} – 4x + 4 = (x – 2)^{2}}\). Solution: \(\color{blue}{x = 2}\) (double root).

Step-by-Step Summary

1. Set the equation equal to zero if not already.
2. Check for a common factor (GCF) and factor it out first.
3. Identify the pattern: standard trinomial, difference of squares, perfect square, or grouping.
4. Write the factored form.
5. Apply the Zero Product Property and solve each factor.
6. Check solutions in the original equation.

Watch: Solving a Quadratic by Factoring

Khan Academy walks through the factoring method for solving quadratics step by step:

Factoring Quadratics – Worked Examples

Example 1: Solve \(\color{blue}{x^{2} + 5x + 6 = 0}\).

Find factors of 6 that add to 5: \(\color{blue}{2 \times 3 = 6}\), \(\color{blue}{2 + 3 = 5}\).
Factor: \(\color{blue}{(x + 2)(x + 3) = 0}\).
Solutions: \(\color{blue}{x = -2}\) or \(\color{blue}{x = -3}\).

Example 2: Solve \(\color{blue}{x^{2} – 7x + 12 = 0}\).

Find factors of 12 that add \(\color{blue}{\text{ to } -7}\): \(\color{blue}{(-3)(-4) = 12}\), \(\color{blue}{(-3)+(-4) = -7}\).
Factor: \(\color{blue}{(x – 3)(x – 4) = 0}\).
Solutions: \(\color{blue}{x = 3}\) or \(\color{blue}{x = 4}\).

Example 3: Solve \(\color{blue}{x^{2} – 25 = 0}\) (difference of squares).

Factor: \(\color{blue}{(x + 5)(x – 5) = 0}\).
Solutions: \(\color{blue}{x = 5}\) or \(\color{blue}{x = -5}\).

Example 4: Solve \(\color{blue}{x^{2} – 4x + 4 = 0}\) (perfect square).

Factor: \(\color{blue}{(x – 2)^{2} = 0}\).
Solution: \(\color{blue}{x = 2}\) (double root).

More Practice: Solving a Quadratic by Factoring

Khan Academy demonstrates an additional factoring example from Algebra I:

Exercises for Factoring Quadratics

1. Factor and solve: \(\color{blue}{x^{2} + 7x + 10 = 0}\)
2. Factor and solve: \(\color{blue}{x^{2} – x – 12 = 0}\)
3. Factor and solve: \(\color{blue}{x^{2} – 36 = 0}\)
4. Factor and solve: \(\color{blue}{x^{2} + 6x + 9 = 0}\)
5. Factor and solve: \(\color{blue}{2x^{2} – x – 6 = 0}\)
6. Factor and solve: \(\color{blue}{3x^{2} + 5x + 2 = 0}\)

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Answers

1. \(\color{blue}{x = -2}\) and \(\color{blue}{x = -5}\)
2. \(\color{blue}{x = 4}\) and \(\color{blue}{x = -3}\)
3. \(\color{blue}{x = 6}\) and \(\color{blue}{x = -6}\)
4. \(\color{blue}{x = -3}\) (double root)
5. \(\color{blue}{x = 2}\) and \(\color{blue}{x = -\frac{3}{2}}\)
6. \(\color{blue}{x = -\frac{2}{3}}\) and \(\color{blue}{x = -1}\)

Free Solving a Quadratic Equation by Factoring Worksheet

Ready to practice on your own? Download our free Solving a Quadratic Equation by Factoring worksheet below, work through each problem at your own pace, and then check your answers. If a few give you trouble, scroll back up to the worked examples and try again — steady practice is the surest way to master Solving a Quadratic Equation by Factoring before a quiz or test.

Download Solving Quadratics by Factoring Worksheet

Frequently Asked Questions

How do I know if a quadratic can be factored?

Check the discriminant: \(\color{blue}{b^{2} – 4\text{ ac }}\). If it is a perfect square (0, 1, 4, 9, 25, …), the quadratic factors over the integers. If not, use the quadratic formula instead.

What is the Zero Product Property?

The Zero Product Property states that if the product of two factors equals zero, then at least one factor must equal zero: \(\color{blue}{\text{ AB } = 0}\) means \(\color{blue}{A = 0}\) or \(\color{blue}{B = 0}\). This is why factoring immediately gives the solutions.

Does factoring work if the equation is not set to zero?

No. You must set the equation equal to zero before factoring. For example, rewrite \(\color{blue}{x^{2} + 3x = 10}\) as \(\color{blue}{x^{2} + 3x – 10 = 0}\) before factoring.

Related Topics

• Solving a Quadratic Equation
• Completing the Square
• Discriminant of a Quadratic Equation
• Quadratic Function
• Graphing Quadratic Functions