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High school🔤Algebra●○○○○· 3 min
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Zero product equation: A × B = 0 in 2 seconds

If a product equals 0, at least one factor is 0. The rule that cracks every factored equation.

The rule

A product of factors is zero if and only if at least one factor is zero.

A×B=0  ⟺  A=0 or B=0A \times B = 0 \iff A = 0 \text{ or } B = 0

Generalization: A×B×C=0  ⟺  A=0A \times B \times C = 0 \iff A = 0 or B=0B = 0 or C=0C = 0.

3-step method

  1. Factor the expression until you get a product equal to 0.
  2. Apply the rule: each factor may be zero.
  3. Solve each simple equation, then collect the solutions.

Typical example

Solve (x−3)(2x+1)=0(x - 3)(2x + 1) = 0.

  • x−3=0⇒x=3x - 3 = 0 \Rightarrow x = 3
  • 2x+1=0⇒x=−1/22x + 1 = 0 \Rightarrow x = -1/2

Solutions: S={−1/2 , 3}S = \{-1/2 \, , \, 3\}.

Example needing factoring

Solve x2−5x=0x^2 - 5x = 0.

Factor: x(x−5)=0x(x - 5) = 0.

  • x=0x = 0
  • x−5=0⇒x=5x - 5 = 0 \Rightarrow x = 5

S={0 , 5}S = \{0 \, , \, 5\}.

ONLY applies when the product equals 0

A×B=12A \times B = 12 does NOT imply that A=12A = 12 or B=12B = 12 (you might have A=3,B=4A = 3, B = 4).

⚠️ Common error: moving 1212 to the left but not factoring. Ex: x(x−1)=6x(x-1) = 6 does not give x=6x = 6 or x−1=6x - 1 = 6.

Common mistakes

  1. Using the rule without a product (no =0=0 or not factored).
  2. Missing a solution (not solving each factor).
  3. Writing x2=25⇒x=5x^2 = 25 \Rightarrow x = 5 forgetting −5-5 (here x2−25=0x^2 - 25 = 0 ⇒ (x−5)(x+5)=0(x-5)(x+5)=0).
?À toi de jouer

Solutions of (2x − 4)(x + 3) = 0?

Mini-quiz

Question 1 / 3

Solutions of (x + 5)(x − 2) = 0?

#equation#zero product#factoring#grade 10

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