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Solving quadratics by factoring: zero on one side, then the zero-product property

If a product is zero, one of the factors is zero. That is the whole method, and it only works when the other side of the equation is 0. The points go when a student factors against a 15, divides both sides by x, or reads x = 5 off the factor (x + 5).

Short answer

Move everything to one side so the equation reads something = 0. Factor that something completely. Set each factor equal to 0 and solve the small equations. Never divide both sides by x; factor the x out instead, so the answer x = 0 survives. A factor of (x + 5) gives x = −5, not 5. Check each answer in the original equation.

Why it works

The zero-product property says: if a · b = 0, then a = 0 or b = 0 (or both). The reason is that if a isn't zero you can divide both sides by it, and b = 0 is what's left. Zero is the only number this works for. a · b = 15 tells you almost nothing: 3 · 5, 1 · 15, 2 · 7.5, (−3) · (−5) all do it. That's why the first step is always to get 0 alone on one side, and why a factored expression next to a 15 can't be solved by setting factors equal to 15.

Factoring turns one quadratic equation into two linear ones, each with a single answer. Two factors, two answers, which is the most a quadratic can have.

Example 1: already equal to zero

Solve x2 − 5x + 6 = 0.

  1. (x − 2)(x − 3) = 0Factor: −2 and −3 multiply to 6 and add to −5.
  2. x − 2 = 0 or x − 3 = 0Zero-product property: one of the two factors must be 0.
  3. x = 2 or x = 3Check: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

Example 2: set it to zero first

Solve x2 + 2x = 15.

  1. x2 + 2x − 15 = 0Subtract 15 from both sides. The zero-product property needs a 0, not a 15.
  2. (x + 5)(x − 3) = 0Factor: 5 and −3 multiply to −15 and add to 2.
  3. x + 5 = 0 or x − 3 = 0Each factor set to 0.
  4. x = −5 or x = 3Check in the original: 25 − 10 = 15 and 9 + 6 = 15. Note the signs: the factor (x + 5) gives −5.

Example 3: factor out the x, don't divide by it

Solve 3x2 = 12x.

  1. 3x2 − 12x = 0Subtract 12x. Everything on one side, 0 on the other.
  2. 3x(x − 4) = 0The greatest common factor is 3x. There's no constant term, so x itself is a factor.
  3. 3x = 0 or x − 4 = 0Each factor set to 0.
  4. x = 0 or x = 4Check: 3 · 0 = 12 · 0 and 3 · 16 = 48 = 12 · 4. Two answers, and 0 is one of them.

Example 4: a leading coefficient that isn't 1

Solve 2x2 + 7x − 4 = 0.

  1. (2x − 1)(x + 4) = 0Factor with the AC method or by trial. Check the middle: 2x · 4 + (−1) · x = 8x − x = 7x.
  2. 2x − 1 = 0 or x + 4 = 0Each factor set to 0.
  3. x = 12 or x = −42x = 1 gives x = 1/2, not 2. Check: 2 · ¼ + 7/2 − 4 = ½ + 7/2 − 4 = 0, and 32 − 28 − 4 = 0.

Common mistakes, and the exact line they happen on

1. Factoring while the other side is 15

Solve x2 + 2x = 15 (Example 2).

  1. x(x + 2) = 15This is the mistake. The factoring is correct, but a product equal to 15 doesn't force either factor to be anything. The student continues with x = 15 or x + 2 = 15, and neither works: 225 + 30 = 255 and 169 + 26 = 195.

Subtract the 15 first. Then the factors are different, (x + 5)(x − 3), and setting them to 0 is legal.

2. Dividing both sides by x

Solve 3x2 = 12x (Example 3).

  1. 3x = 12This is the mistake. Both sides were divided by x. That's only allowed if x ≠ 0, so the step silently assumes 0 isn't an answer, and it is. The student gets x = 4 and loses x = 0.

Dividing by a variable throws away the solution where that variable is zero, and no later check brings it back. Move the term over and factor the x out instead.

3. Reading the root with the wrong sign

  1. (x + 5)(x − 3) = 0Correct so far.
  2. x = 5 or x = −3This is the mistake. The signs were copied straight out of the parentheses. x + 5 = 0 means x = −5. Check x = 5: 25 + 10 − 15 = 20, not 0.

Write the little equation x + 5 = 0 on its own line before solving it. That one extra line removes the flip.

4. Forgetting the negative root

  1. x2 = 9Start.
  2. x = 3This is the mistake. (−3)² is also 9. Half the answer is missing.

Treat it like every other quadratic: x2 − 9 = 0, then (x − 3)(x + 3) = 0, a difference of squares, so x = 3 or x = −3. The factored form produces both roots automatically.

Practice

  1. Solve x2 + 8x + 12 = 0.

    Show answer

    (x + 2)(x + 6) = 0, so x = −2 or x = −6.

  2. Solve x2 = 7x.

    Show answer

    x2 − 7x = 0, x(x − 7) = 0, so x = 0 or x = 7. Don't divide by x.

  3. Solve x2 − 3x = 10.

    Show answer

    x2 − 3x − 10 = 0, (x − 5)(x + 2) = 0, so x = 5 or x = −2.

  4. Solve 4x2 − 25 = 0.

    Show answer

    (2x − 5)(2x + 5) = 0, so x = 52 or x = −52.

  5. Solve 3x2 + 10x − 8 = 0.

    Show answer

    (3x − 2)(x + 4) = 0 (check the middle: 12x − 2x = 10x), so x = 23 or x = −4.

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