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Zero Product Property

Zero Product Property – definition and example, Algebra 1 math glossary – Mathovore

Key takeaway: The zero product property states that if two or more factors multiply to give 0, then at least one of those factors must itself equal 0.

Grade 9 – Algebra 1A-SSE.B.3aExpressions and Equations

Definition of Zero Product Property

Definition

The zero product property explains why factoring can solve equations. Zero is special: no two nonzero numbers multiply to make zero. So if \(A \cdot B = 0\), then either \(A = 0\), or \(B = 0\), or both.

To use it, first arrange the equation so that its right side is 0, then factor the left side. Turn each factor into its own mini-equation with 0 on the right, and handle them one at a time. Each solution makes one factor vanish, which drags the entire product down to zero. The property applies to any number of factors, but only when the product equals exactly zero.

Formula and notation

In symbols, the zero product property says:

\[A \cdot B = 0 \ \Longrightarrow\ A = 0 \ \text{ or } \ B = 0\]

For a factored equation such as \((px + q)(rx + s) = 0\), solve \(px + q = 0\) and \(rx + s = 0\). The two answers are the zeros of the quadratic.

(2x − 9)(x + 4) = 02x − 9 = 0x = 9/2x + 4 = 0x = −4
A product equal to 0 splits into two simple equations, one per factor.

Key facts

  • The zero product property needs a product of exactly 0; a product of 6 or 1 gives no such rule.
  • Each factor gives its own equation and its own solution.
  • A repeated factor, like (x − 4)², gives one solution.
  • Plug each answer back in to confirm it works.

Zero Product Property example

Problem

Solve \((3x + 7)(x – 5) = 0\).

  1. The product is 0, so one of the factors must be 0.

  2. First factor: \(3x + 7 = 0\), so \(3x = -7\) and \(x = -\frac{7}{3}\).

  3. Second factor: \(x – 5 = 0\), so \(x = 5\). Check \(x = 5\): \((15 + 7)(0) = 0\).

The roots are \(x = -\frac{7}{3}\) and \(x = 5\).

Common mistake

Watch out

Given \((x + 2)(x – 5) = 18\), Leo sets \(x + 2 = 18\) and \(x – 5 = 18\), and gets \(x = 16\) and \(x = 23\).

Why it is wrong

The splitting rule needs a product of 0. Many pairs of numbers multiply to 18, so a factor need not equal 18. Expand to \(x^{2} – 3x – 10 = 18\), subtract 18 on both sides, leaving \(x^{2} – 3x – 28 = 0\), factor as \((x – 7)(x + 4) = 0\), and only then split: \(x = 7\) or \(x = -4\).

Related terms and lessons

Browse the glossary: A–Z list · Algebra 1 vocabulary · Expressions and Equations

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