How do I solve a quadratic equation by factoring?
Quick answer
First, rewrite the quadratic so one side is zero. For a monic trinomial x² + bx + c, find two numbers that multiply to c and add to b, then use them to write two factors. Set each factor equal to zero and solve. Finally, substitute each candidate into the original equation to check that it makes the equation true. [S1]
The question, the details, the next step.
What does solving a quadratic by factoring involve?
You rewrite the quadratic as a product of simpler factors, then find the values of x that make that product zero. The key is to get the equation equal to zero first. For a monic trinomial—one whose x² coefficient is 1—look for two numbers whose product is the constant term and whose sum is the coefficient of x. Those numbers help form the factors. [S1]
For example, consider x² + 5x + 6 = 0. We need two numbers that multiply to 6 and add to 5. They are 2 and 3, so the expression factors as:
(x + 2)(x + 3) = 0
Factoring rewrites the expression without changing which x-values make the equation true.
How do I get from the factors to the answers?
Use the zero-product property: if a product equals zero, at least one of its factors must equal zero. So, for (x + 2)(x + 3) = 0, set each factor to zero in turn: [S1]
x + 2 = 0, sox = -2.x + 3 = 0, sox = -3.
The candidate solutions are -2 and -3. Notice the signs: solving x + 2 = 0 gives -2, not 2. This method applies after the equation has been written as a product equal to zero.
Can you check both answers in the original equation?
Yes. Substitute each value into the original equation, x² + 5x + 6 = 0. A candidate is a solution if the left-hand side evaluates to zero. [S1]
For x = -2:
(-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0
For x = -3:
(-3)² + 5(-3) + 6 = 9 - 15 + 6 = 0
Both checks give zero, so both values solve this equation. Checking in the original equation can catch a factoring or sign error.
What if I can’t find the two numbers, or my equation isn’t equal to zero yet?
First, move all terms to one side so the other side is zero, and combine like terms. Then check whether the result is a monic trinomial. For that form, the number pair must meet both tests: its product is the constant term, and its sum is the x coefficient. [S1]
For example, in x² + 7x + 12 = 0, the pair 3 and 4 works: 3 × 4 = 12 and 3 + 4 = 7. So the equation becomes (x + 3)(x + 4) = 0, giving x = -3 or x = -4; substitution can check each value. Don’t force a pair that fails either test. The number-pair steps described here cover monic trinomials; they do not establish that every quadratic factors neatly using this method.
A practical checklist
Rewrite the equation with zero on one side and combine like terms.
For a monic trinomial, find a number pair whose product is the constant and whose sum is the x coefficient.
Write the two factors and check that they match the trinomial.
Set each factor equal to zero and solve for x.
Substitute each candidate into the original equation and confirm it evaluates to zero.
A few more useful answers
Why must one side of the equation be zero before I set the factors to zero?
The zero-product property applies when a product equals zero. Put the quadratic in the form of a product equal to zero before setting each factor to zero. [S1]
How can I check whether my proposed answers are solutions?
Substitute each candidate into the original equation. It is a solution when the equation evaluates to zero. [S1]
What if no number pair fits the monic trinomial?
Do not claim a pair works if it fails the sum or product test. The number-pair steps described here cover monic trinomials; they do not show that every quadratic can be factored this way. [S1]
Sources
References used for this guide.
- [S1] OpenStax College Algebra: The Zero-Product Property and Solving Quadratics by Factoring openstax.org · Checked