# Lesson 7

These materials, when encountered before Algebra 1, Unit 7, Lesson 7 support success in that lesson.

## 7.1: Math Talk: Missing Values (5 minutes)

### Warm-up

The purpose of this Math Talk is to elicit strategies and understandings students have for finding a value that will multiply to a certain number. These understandings help students develop fluency and will be helpful in the associated Algebra 1 lesson when students will need to be able to find missing values to help factor quadratic expressions.

### Launch

Display one problem at a time. 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.

### Student Facing

Mentally solve each equation for $$a$$.

$$7\boldcdot a = 49$$

$$7 \boldcdot a= \text{-}49$$

$$\text{-}7\boldcdot a= 49$$

$$\text{-}7\boldcdot a = \text{-}49x$$

### Activity Synthesis

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?”

## 7.2: Finding Pairs that Work (15 minutes)

### Activity

In this activity, students find pairs of integers that have a given product and given sum. In the associated Algebra 1 lesson, students factor quadratic expressions in standard form that have a negative constant term. Finding pairs of values with a given negative product and given sum will help students find values to try while factoring.

### Student Facing

For each question, find a pair of integers with the given product and sum.

1. product: 6, sum: 5
2. product: 6, sum: 7
3. product: 4, sum: -5
4. product: -1, sum: 0
5. product: -6, sum: 1
6. product: -12, sum: -1
7. product: -12, sum: 4

### Activity Synthesis

The purpose of the discussion is to ensure students have some strategies for finding a pair of integers that have a given product and sum. Select students to share their solutions and any strategies they have to finding the values. Ask students to complete the table with information they know about the signs of the integers.

positive product negative product
positive sum

larger integer

smaller integer

larger integer

smaller integer

negative sum

larger integer

smaller integer

larger integer

smaller integer

## 7.3: Factor Expansion (20 minutes)

### Activity

In this activity, pairs of students expand binomial factors into the standard form of a quadratic expression. Each student then compares their original question and solution with their partner to notice what is the same and different. In particular, students should notice how rearranging factors and changing the operation within factors can change the standard form. In the associated Algebra 1 lesson, students factor quadratic expressions involving a negative constant term. The fluency and pattern recognition from this activity should help students recognize similar patterns when factoring. Students have the opportunity to construct viable arguments and critique the reasoning of others (MP3) when they work with a partner to ensure two different problems give the same solution.

### Launch

Arrange students in groups of 2. Help students understand the instructions by reading them together and asking for any clarifying questions.

### Student Facing

For each question:

• rewrite the expression in standard form
• be prepared to explain anything you notice in the comparison

Partner A:

1. $$(x-1)(x-2)$$
2. $$(x-1)(x+2)$$
3. $$(x+4)(x-4)$$
4. $$(x+3)(x-6)$$
5. $$(x-2)(x-3)$$
6. $$(x-2)(x+7)$$
7. $$(x+5)(x-2)$$
8. $$(4-x)(1-x)$$

Partner B:

1. $$(x+1)(x+2)$$
2. $$(x+1)(x-2)$$
3. $$(x-4)(x+4)$$
4. $$(x-3)(x+6)$$
5. $$(2-x)(x-3)$$
6. $$(x+7)(x-2)$$
7. $$(x-5)(x+2)$$
8. $$(x-4)(x-1)$$

### Student Response

• For each term in a factor involving subtraction that is switched (questions 5 and 8), the terms in the standard form have opposite signs. This is because $$\text{-}(x-a) = (a-x)$$.