Materials & Prep
Prepare a board or projected display with the equations in the lesson and a five-column organizer: original equation, factored equation, zero-product equations, solutions, and meaning. No materials beyond ordinary classroom setup are needed.
Opening prediction
Display two multiplication statements: 8 × 0 = 0 and 8 × 3 = 24. Ask students to predict: “If two numbers multiply to make 0, what must be true about at least one of the numbers?” Students answer independently, then compare with a partner. Confirm that at least one factor must equal 0. Connect: “Today we will use that familiar fact to solve equations when the factors contain x. The key question will be: what values make the original equation equal to zero?”
Model the reason
ConnectA product is zero when at least one factor is zero. If x = 2, the first factor is zero. If x = 3, the second factor is zero. This is the zero-product property, and it applies when a product equals zero, not when it equals an arbitrary constant.
Worked example
x² - 5x + 6 = 0 (x - 2)(x - 3) = 0 x - 2 = 0 or x - 3 = 0 x = 2 or x = 3
Think aloudI first factor the left side, but I do not change the right side. The equation still says the product equals zero. I then set each factor equal to zero because either factor can make the product zero. The common error is to use the constant term 6 as the new right side after factoring. That would change the equation.
ConfrontShow the incorrect route: (x - 2)(x - 3) = 6. Expanding gives x² - 5x + 6 = 6, so x² - 5x = 0 and x(x - 5) = 0, producing x = 0 or x = 5. Check both in the original equation: at x = 0, the left side is 6, and at x = 5, the left side is 6. Neither makes the original equation equal zero. The correct solutions are x = 2 and x = 3.
Interpret: For y = x² - 5x + 6, the solutions x = 2 and x = 3 are the x-values where y = 0. The graph crosses the x-axis at (2, 0) and (3, 0). A solution is not just a number from a factoring procedure; it identifies an input that makes the original quadratic equal zero.
Guided practice
ModelComplete the first equation with students, requiring them to say what the right side remains after factoring.
x² + x - 12 = 0 (x + 4)(x - 3) = 0 x + 4 = 0 or x - 3 = 0 x = -4 or x = 3
AskWhy is the equation still set equal to zero? What do the solutions mean if this quadratic is written as y = x² + x - 12? Expected response: They are the x-values where y equals zero, so the x-intercepts are (-4, 0) and (3, 0).
Check for understandingStudents solve 2x² - 10x = 0 on their own boards. Expected work is 2x(x - 5) = 0, so 2x = 0 or x - 5 = 0, giving x = 0 or x = 5. Students must circle the zero on the right side and write one sentence explaining why the product-zero step is valid. If a student writes the constant term as the right side, have the student substitute that proposed answer into the original equation before revising it.
ScaffoldDirect students through the organizer one column at a time. Provide the sentence frame, “The factored product equals zero, so I set equal to zero or equal to zero. The solutions mean .”
Independent application
Have students solve and interpret each problem. For each, they must show factoring, use the zero-product property, and state what the solutions mean.
- Solve x² + 2x - 15 = 0. State the x-intercepts if the equation is y = x² + 2x - 15.
- Solve 2x² - 10x = 0. Explain why the factor 2 does not create an additional solution.
- A ball’s height above the ground is h(t) = -t² + 6t, where t is time in seconds. Find when h(t) = 0 and interpret both solutions in this situation.
CirculateLook for students who factor correctly but write the factored expression equal to the constant term. Ask, “What is the right side of the original equation, and did factoring change it?” Also ask students to verify at least one solution in the original equation.
ExpectedFor problem 1, (x + 5)(x - 3) = 0, so x = -5 or x = 3. The x-intercepts are (-5, 0) and (3, 0). For problem 2, 2x(x - 5) = 0, so x = 0 or x = 5. The factor 2 is never zero, so it adds no solution. For problem 3, -t(t - 6) = 0, so t = 0 or t = 6. The ball is at ground level at 0 seconds and 6 seconds. The negative-time solution is not relevant because time in this situation begins at 0 seconds.
ExtensionAsk students to explain why the equation 2x(x - 5) = 0 has the same solutions as x(x - 5) = 0, and to describe a situation in which one algebraic solution would need to be rejected because of the meaning of the variable.
Closing check
Exit ticketSolve x² - 7x + 12 = 0 and answer both questions: What are the solutions? What do they mean for y = x² - 7x + 12? Students must show the factored equation and the two zero-product equations.
Expected(x - 3)(x - 4) = 0, so x - 3 = 0 or x - 4 = 0, giving x = 3 or x = 4. These are the x-values where y = 0, so the x-intercepts are (3, 0) and (4, 0).
Ask students to connect the opening fact to today's solutions in one sentence: “Both values work because .” Collect the exit ticket and revision to check whether students used the zero-product property and interpreted the solutions rather than only listing numbers.