Lesson 4

This Math Talk introduces students to the zero product property and prepares them to use it to solve quadratic equations. It reminds students that if two numbers are multiplied and the result is 0, then one of the numbers has to be 0. Answering the questions mentally prompts students to notice and make use of structure (MP7).

Display one problem at a time and ask students to respond without writing anything down. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

• “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
• “Do you agree or disagree? Why?”

Highlight explanations that state that any number multiplied by 0 is 0. Then, introduce the zero product property, which states that if the product of two numbers is 0, then at least one of the numbers is 0.

In this activity, students solve equations of increasing complexity and do so by reasoning. They begin with linear equations, move toward a series of quadratic expressions in factored form, and end with a cubic expression in factored form. The progression prompts students to reason about the parts and structure of the expressions (MP7), rather than to memorize steps for solving without understanding, and to notice regularity through repeated reasoning (MP8).

As students discuss their reasoning with their partner, listen for those who invoke the zero product property to explain how the last four equations could be solved, and those who notice a pattern in how the equations could be solved. (Though the last question involves a cubic equation, solving it involves the same reasoning as solving quadratic expressions.)

Arrange students in groups of 2. Tell students to work quietly and answer at least half of the questions before discussing their thinking with a partner.

If needed, remind students that some equations have more than one solution. Because we want students to use reasoning and the structure of equations to develop their solutions, discourage use of graphing technology or spreadsheets in this activity.

Students may incorrectly think that \(x\) can represent a different value in each factor in an equation. For example, upon finding -11 and 3 as solutions to \((x-3)(x+11)=0\), they think that one solution is for the \(x\) in \((x-3)\) and the other for the \(x\) in \((x-11)\).

Remind students that solving the equation \((x-3)(x+11)=0\) is like finding the zeros of the function defined by \((x-3)(x-11)\). Although there may be two values of \(x\) that lead to 0 for the value of\((x-3)(x-11)\), only one input can be entered into the function at a time. Ask students to substitute the solutions into the equations and check if the expression is equal to 0 each time.

• When \(x=\text-11\), the value of the expression is \((\text-11-3)(\text-11 +11)\) or \((\text-14)(0)\), which is 0.
• When \(x=3\), the value of the expression is \((3-3)(3+11)\) or \((0)(14)\), which is 0.

Invite students to share their strategies for solving the non-linear equations. As they explain, record and organize each step of their reasoning process and display for all to see.

For example, the equation \((x-3)(2x+11)=0\) tells us that, if the product of \((x-3)\) and \((2x+11)\) is 0, then either \(x-3\) is equal to 0, or \(2x+11\) is equal to 0. We can then organize the rest of the solving process as:

The equation is true when \(x = 3\) and when \(x=\text- \frac{11}{2}\).

Emphasize that because at least one of the factors must be 0 for the product to be 0, we can write each expression that is a factor to equal to 0 and solve each of these equations separately.

Remind students that we can check our solutions by substituting each one back into the equation and see if the equation remains true. Although the two factors, \((x-3)\) and \((2x+11)\), won’t be 0 simultaneously when 3 or \(\text- \frac{11}{2}\) is substituted for \(x\), the expression on the left side of the equation will have a value of 0 because one of the factors is 0.

• When \(x\) is 3, the expression is \(\left(3-3\right)\left(2(3)+11\right)\) or \((0)(17)\), which is 0.
• When \(x\) is \(\text-\frac{11}{2}\), the expression is \(\left(\text-\frac{11}{2}-3\right)\left(2(\text-\frac{11}{2} ) +11\right)\) or \(\left(\text-\frac{17}{2}\right)\left(0\right)\), which is 0.

This activity enables students to apply the zero product property to solve a contextual problem and reinforces the idea of solving quadratic equations as a way to reason about quadratic functions.

Previously, students have encountered two equivalent quadratic expressions that define the same quadratic function. Here, they work to show that two quadratic expressions—one in standard form and the other in factored form—really do define the same function. There are several ways to do this, but an efficient and definitive way to show equivalence would be to use the distributive property to expand quadratic expressions in factored form.

Next, they consider which of the two forms helps them find the zeros of the function and then use it to find the zeros without graphing. The work here reiterates the connections between finding the zeros of a quadratic function and solving a quadratic equation where a quadratic expression that defines a function has a value of 0.

Keep students in groups of 2. Prepare access to graphing technology and spreadsheet tool, in case requested.

Display the two equations that define \(h\) for all to see. Tell students that the two equations define the same function. Ask students how they could show that the two equations indeed define the same function.

Give students a moment of quiet time to think of a strategy and test it, and then time to discuss with a partner, if possible. Then, discuss their responses. Some likely strategies:

• Graph both equations on the same coordinate plane and show that they coincide.
• Inspect a table of values of both equations and show that the same output results for any input.
• Use the distributive property to multiply the expression in factored form to show that \((\text-5t-3)(t-6) =\text-5t^2+27t+18\). (Only this reasoning is really a “proof,” but the other methods supply a lot of evidence that they are the same function.)

Once students see some evidence, ask students to proceed to the activity.

Ask students to share their responses and reasoning. Discuss questions such as:

• “Why is the factored form more helpful for finding the time when the object has a height of 0 meters?” (To find the input values when the output has a value of 0 is to solve the equation \(\text {quadratic expression}=0\). When the expression is in factored form, we can use the zero product property to find the unknown inputs.)
• “What if we tried to solve the equation in standard form by performing the same operation to each side?” (We would get stuck. For instance, we could add or subtract terms from each side, but then there are no like terms to combine on either side, so we are no closer to isolating the variable.)

If no students related solving equations in factored form to using the factored form to find the horizontal intercepts of a graph of a quadratic function, discuss that connection.

• “In an earlier unit, we saw that the factored form of a quadratic expression such as \((x-5)(x+9)\) allows us to see the \(x\)-intercepts of its graph, but we didn’t look into why the graph crosses the \(x\)-axis at those points. Can you explain why it does now?” (The \(x\)-intercepts have a \(y\)-value of 0, which means the quadratic function is 0 at those \(x\)-values: \((x-5)(x+9)=0\). If multiplying two numbers gives 0, one of them must be 0. So either \(x-5=0\) or \(x+9 =0\). If \(x-5=0\), then \(x\) is 5. If \(x+9 =0\), then \(x\) is -9.)