Solving Quadratic Equations by Factoring

Factoring can turn a quadratic equation into two simple linear equations. The method rests on one rule: if a product equals zero, at least one factor must equal zero. The order of the steps matters. A quadratic must be written with zero on one side before its factors can be set equal to zero. Factoring an expression is useful, but setting factors equal to zero before the equation is balanced at zero can give an invalid result.

The zero-product property

If ab = 0, then a = 0 or b = 0 (or both). For example, (x − 3)(x + 2) = 0 is true when x − 3 = 0 or x + 2 = 0. The solutions are x = 3 and x = −2. These values are the roots or zeros of the quadratic. They make the product, and therefore the original quadratic expression, equal zero.

This property does not say that if a product equals a nonzero number, one factor must equal zero. For example, (x − 3)(x + 2) = 5 cannot be split into x − 3 = 5 or x + 2 = 5. First transform the full equation so one side is zero; then factor.

Step 1: put the equation in standard zero form

A quadratic equation has a term involving x², and it is commonly written ax² + bx + c = 0 with a ≠ 0. If the equation is x² + 5x = 14, subtract 14 from both sides to get x² + 5x − 14 = 0. Only after moving all terms to one side should you look for factors.

Choose a consistent side and preserve signs while moving terms. If you start with 2x² = 7x + 4, subtract 7x and 4 from both sides: 2x² − 7x − 4 = 0. A sign error in this rearrangement changes the roots, so it is worth checking that the new equation is equivalent to the given one before factoring.

Step 2: factor the quadratic

When the leading coefficient is 1, look for two numbers whose product is c and whose sum is b. For x² + 5x − 14, the numbers 7 and −2 multiply to −14 and add to 5. Therefore x² + 5x − 14 = (x + 7)(x − 2). Always multiply the factors back out to confirm they reproduce all three terms.

When a is not 1, look for a factorization that accounts for the leading coefficient too. For 2x² − 7x − 4, one factorization is (2x + 1)(x − 4), because multiplication gives 2x² − 8x + x − 4, which simplifies to 2x² − 7x − 4. If a factorization is not obvious, a systematic method such as splitting the middle term can help. Not every quadratic with integer coefficients factors neatly over the integers.

Step 3: set each factor equal to zero

Once x² + 5x − 14 = 0 is rewritten as (x + 7)(x − 2) = 0, apply the zero-product property: x + 7 = 0 or x − 2 = 0. Solve each equation separately to get x = −7 or x = 2. There are two solutions because a quadratic can have two real roots, although some quadratics have one repeated real root or no real roots.

For (2x + 1)(x − 4) = 0, the first equation gives 2x + 1 = 0, so x = −1/2; the second gives x − 4 = 0, so x = 4. Divide by the full coefficient of x in each factor. Do not drop a negative sign or assume every solution is an integer.

A quadratic with a common factor

Look for a greatest common factor before applying other factoring patterns. Solve 3x² − 12x = 0 by factoring out 3x: 3x(x − 4) = 0. The factors give x = 0 or x − 4 = 0, so x = 0 or x = 4. Dividing both sides by x at the start would lose x = 0, one of the valid solutions. Factoring preserves it.

A perfect-square pattern and repeated roots

Some trinomials factor as a square of a binomial. For x² − 6x + 9 = 0, the expression is (x − 3)² = 0. The repeated factor gives x − 3 = 0, so x = 3. This is one distinct real solution with multiplicity two. On a graph, the parabola touches the x-axis at that root rather than crossing it.

Worked example from start to finish

Solve x² − 4x = 12. First subtract 12 from both sides: x² − 4x − 12 = 0. Find two numbers with product −12 and sum −4: −6 and 2. Factor to get (x − 6)(x + 2) = 0. Set each factor to zero: x − 6 = 0 gives x = 6, and x + 2 = 0 gives x = −2. Check: 6² − 4(6) = 36 − 24 = 12; (−2)² − 4(−2) = 4 + 8 = 12. Both satisfy the original equation.

What if the quadratic does not factor easily?

Factoring over the integers is not always available. For example, x² + x + 1 has no pair of integers whose product is 1 and whose sum is 1. In that case, another method may be needed, such as completing the square or the quadratic formula, depending on the course and question. Do not force a pair of factors that does not work. Expanding a proposed factorization is the fastest way to test it.

The quadratic formula also reveals why there can be zero, one, or two real roots, but factoring questions often expect recognition of a simple product. If the coefficients are small integers, check for a greatest common factor, a difference of squares, a perfect-square trinomial, or a pair of factors whose product and sum match. If those checks fail, choose another method rather than spending time guessing.

Common mistakes

• Factoring before making one side zero, then setting factors equal to zero anyway.
• Using the wrong signs in a pair of factors. The product must match the constant term and the sum must match the middle coefficient when the leading coefficient is 1.
• Dividing by a variable and losing a possible zero solution.
• Stopping after finding one root even when the equation factors into two different linear factors.
• Accepting an unverified factorization. Multiply it back out.
• Failing to check the answers in the original equation after moving terms.

Exam takeaway

Use the sequence: rewrite as ax² + bx + c = 0, factor, apply the zero-product property, solve every factor equation, and check. The key idea is that a product can equal zero only when a factor equals zero. If the quadratic does not factor cleanly, recognize that the factoring method may not be the right tool for that particular expression.