Solving by factoring

If (x − 2)(x + 3) = 0, then x − 2 = 0 or x + 3 = 0, so the solutions are 2 and −3. No other number makes the product zero.

Dividing both sides by x loses a root

Here is a tempting move. To solve x² = 5x, divide both sides by x and get x = 5. It checks: 25 = 25. But x = 0 works too, since 0 = 0, and it has vanished. Division by x is allowed only when x is not 0, and 0 was exactly the solution that got thrown away. The safe route keeps everything on one side: x² − 5x = 0, so x(x − 5) = 0, and both 0 and 5 come out. The rest of this page is about that route.

If a product is 0, one of the factors is 0

A quadratic equation is an equation with a second-degree polynomial, usually written in standard form ax² + bx + c = 0. Often the easiest way to solve one is factoring. It rests on a single fact, the zero-product property: if a · b = 0, then a = 0 or b = 0. The reason is that multiplying anything by zero gives zero, and two nonzero numbers never multiply to zero. So if the left side can be written as a product of two simpler factors, the hard question splits into two easy ones.

Factoring is multiplication run backwards. OpenStax shows it with (x − 2)(x + 3): multiply out and you get x² + 3x − 2x − 6 = x² + x − 6. So the equation x² + x − 6 = 0 can be rewritten as (x − 2)(x + 3) = 0. By the zero-product property, x − 2 = 0 or x + 3 = 0, which gives x = 2 or x = −3. Check both: 4 + 2 − 6 = 0 and 9 − 3 − 6 = 0. On the graph of y = x² + x − 6 these are the two x-intercepts.

Product 6, sum −5: finding the pair for x² − 5x + 6

For x² + bx + c, look for two numbers that multiply to c and add to b. For x² − 5x + 6 you need a product of 6 and a sum of −5. The pairs that multiply to 6 are 1 and 6, 2 and 3, and their negatives; only −2 and −3 add to −5. So x² − 5x + 6 = (x − 2)(x − 3), and the solutions are 2 and 3. The signs give hints: a positive c means both numbers have the same sign, the sign of b; a negative c means one is positive and one is negative.

Two special shapes are worth spotting. First, a greatest common factor: 3x² − 12x = 0 has 3x in both terms, so 3x(x − 4) = 0 and x = 0 or x = 4. Do not divide both sides by x. As at the top of the page, that throws away the solution 0. Second, a difference of squares: x² − 4 = (x − 2)(x + 2), so x² − 4 = 0 gives x = 2 or x = −2. The same equation can also be solved with the square root property: x² = 4, so x = ±2.

When a is not 1 the search takes a little longer. For 2x² + 5x − 3 = 0, try factors of the form (2x + p)(x + q) with pq = −3. The pair p = −1, q = 3 works: (2x − 1)(x + 3) = 2x² + 6x − x − 3 = 2x² + 5x − 3. Setting each factor to zero gives x = 1/2 and x = −3. If no pair of whole numbers works, the quadratic does not factor nicely, and the quadratic formula in the next unit still solves it.

A 40 m² rectangle, and the root you throw away

A rectangle is 3 meters longer than it is wide and its area is 40 square meters. Call the width w; the length is w + 3, so w(w + 3) = 40, which becomes w² + 3w − 40 = 0. Two numbers with product −40 and sum 3 are 8 and −5, so (w + 8)(w − 5) = 0 and w = −8 or w = 5. A width cannot be negative, so the width is 5 m and the length 8 m. Check: 5 × 8 = 40. Rejecting a solution that does not fit the situation is a normal step.