# Study Guide

## Field 242: Multi-Subject: Secondary Teachers (Grade 7– to Grade 12) Part Two: Mathematics

### Sample Constructed-Response Item

#### Competency 0004 Analysis, Synthesis, and Application

start bold Use the data provided to complete the task that follows.  end bold

Using the data provided, prepare a response of approximately 400 to 600 words in which you:

• identify a significant mathematical strength related to the given standard that is demonstrated by the student, citing specific evidence from the exhibits to support your assessment;
• identify a significant area of need related to the given standard that is demonstrated by the student, citing specific evidence from the exhibits to support your assessment; and
• describe an instructional intervention that builds on the student's strengths and that would help the student improve in the identified area of need. Include a strategy for helping the student build a viable argument related to the given standard.

#### Background Information

A teacher is working with a student in a high-school algebra class. The class is currently working on the following standard from the New York State P–12 Common Core Learning Standards for Mathematics.

start bold Reasoning with Equations & Inequalities (A-REI) end bold

start bold Solve equations and inequalities in one variable. end bold

4b. Solve quadratic equations in one variable. Solve quadratic equations by inspection  left paren e.g., for x squared equals 49 right paren  taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as  a plus or minus B I   for real numbers a and b.

In particular, the class is working on solving quadratic equations by factoring. Previous classes have covered the structure of expressions, writing expressions in equivalent form, and performing arithmetic with polynomials.

#### Student Work Sample

The student was asked to solve several quadratic equations by factoring. Shown below are two sample problems along with the student's response to each problem. Both the problems and the responses are representative of the student's work.

Worksheet Directions: Solve for x. Check your work and justify the steps you used to find the solution.

Problem 1:  x squared plus 2x minus 3 equals zero

 x2 + 2x − 3 = 0 x squared plus 2 x minus 3 equals 0  factor this (x )(x ) x times x  look at –3 look at negative 3  – 3 • 1 = –3       3 • –1 = –3 negative 3 times 1 equals negative 3, 3 times negative 1 equals negative 3  3 + –1 = 2 3 plus negative 1 equals 2  (x + 3)(x − 1) = 0 left paren x plus 3 right paren, left paren x minus 1 right paren equals 0  solve each of these for x (x + 3) left paren x plus 3 right paren  x + 3 = 0 x plus 3 equals 0  −3 −3 minus 3, minus 3  x = –3 x equals negative 3  (x − 1) left paren x minus 1 right paren  (x − 1) = 0 left paren x minus 1 right paren equals 0  x − 1 = 0 x minus 1 equals 0  +1 +1plus 1, plus 1 x = +1 x equals plus 1 (x + 3)(x − 1) = x • x + x • –1 left paren x plus 3 right paren left paren x minus 1 right paren equals x times x plus x times negative 1  + 3x + 3 • –1 plus 3 x plus 3 times negative 1  x2 − x + 3x − 3 x squared minus x plus 3 x minus 3  check answer x = –3 x equals negative 3  x2 + 2x − 3 = 0 x squared plus 2 x minus 3 equals 0  (–3)2 + 2(–3) − 3 left paren negative 3 right paren squared plus 2 left paren negative 3 right paren minus 3  9 − 6 − 3 9 minus 6 minus 3  0 = 0 0 equals 0, checked  x2 + 2x − 3 = 0 x squared plus 2 x minus 3 equals 0  (+1)2 + 2(+1) − 3 = 0 left paren positive 1 right paren squared plus 2 left paren positive 1 right paren minus 3 equals 0  1 + 2 − 3 = 0 1 plus 2 minus 3 equals 0  0 = 0 0 equals 0, checked

You find the solution by solving the equation. Plug the answer back into the equation to check it.

Problem 2:  x squared plus 6 x equals negative 8

 to factor this take out an x x2 + 6x = –8 x squared plus 6 x equals negative 8  Now you can solve it. (x)(x + 6) = –8 left paren x right paren left paren x plus 6 right paren equals negative 8  x = –8 x equals negative 8  (x + 6) left paren x plus 6 right paren  x + 6 = –8 x plus 6 equals negative 8  6 = 6 6 equals 6  x = –2 x equals negative 2 check x2 + 6x = 8 x squared plus 6 x equals 8  (8)2 + 6(8) left paren 8 right paren squared plus 6 left paren 8 right paren   64 + 48 64 plus 48  112 x2 + 6x x squared plus 6 x  (–2)2 + 6(–2) left paren negative 2 right paren squared plus 6 left paren negative 2 right paren  4 + 12 4 plus 12  16

Since this is an equation, both sides must be equal, so you set them equal to each other and solve for x.

#### Excerpt of Interview with Student

Shown below is an excerpt of an interview between the teacher and the student in which the teacher questions the student about the method of solution used to solve the problems shown in the student work sample.

 Teacher: We will start with Problem 1. Can you tell me what you did to solve this equation? Student: First I wrote down two sets of parentheses. Then I factored it by looking for two numbers that when multiplied give 3, no, minus 3, and when you add them you get 2. I wrote out possible answers and found that 3 and minus 1 worked. Then I put them in with the x's like here (points to the product of two binomials set equal to zero). Teacher: It looks like after you factored the trinomial into two binomials, you set each of them equal to zero and solved the resulting equation. Can you tell me why you took that step? Can you justify why you set each binomial equal to zero? Student: Whatever is on the left side of the equation has to equal what's on the right side. So (x + 3)  left paren x plus 3 right paren  has to equal zero and so does (x − 1),  left paren x minus 1 right paren,  because equations have to balance. Then you just add or subtract the number to find the answer. Teacher: Let's look at Problem 2. Can you tell what strategy you used to solve the equation? Student: I solved it just like I solved the first one. This time, though, it was easier to solve because it's easier to factor. You just take out an x. Teacher: After you factored the left side of the equation, what did you do next? Student: I set both sides of the equation equal to 8 and solved for x. Teacher: Can you justify why you took that step? Student: Both sides of an equation must be equal. So whatever is on the left side has to equal what is on the right side. So I set them equal to each other and solved for x.

### Sample Strong Response to the Constructed-Response Assignment

A significant strength demonstrated by the student is her understanding of how to factor a polynomial. When factoring Problem 1, the student knew two numbers were needed that "when multiplied give minus 3, and when you add them you get 2" as the constant terms in the binomials. Although she made many errors in Problem 2, she correctly factored  x squared plus 6 x equals negative 8 as left paren x right paren left paren x plus 6 right paren equals 8.

One significant need demonstrated by the student is her misunderstanding of the Zero Product Property. A zero product indicates that at least one of the factors must also equal zero. The student demonstrates an incomplete understanding of this property in the interview when she explains that both  left paren x plus 3 right paren  and  left paren x minus 1 right paren   have to equal zero "because equations have to balance." She then apparently generalizes this flawed understanding and incorrectly concludes that a product other than zero must have factors equal to that number, as shown in Problem 2:  x equals negative 8  and  x plus 6 equals negative 8  and her explanation that "I set both sides of the equation equal to 8" because "both sides of an equation must be equal." The student may not know the reason for factoring, and is simply factoring the polynomials because the assignment was "solve several quadratic equations by factoring."

Instructional intervention should start with the Zero Product Property. This concept is pivotal in understanding how and why to use factoring as a solution method for quadratic equations. The teacher could ask her to identify a list of possible solutions for A and B in the problem  A times B equals 0  Then ask for a list of possible solutions for A and B in the problem  A times B equals 12  The student will then compare the lists. The teacher should question the student to help her understand the property: What does she notice? Which list is easier to create and why? What is special about the number 0 when you are finding factors?

Next, ask the student to solve a problem such as this:  left paren x plus 8 right paren left paren x minus 2 right paren equals 0  The student will probably solve this correctly, as she correctly solved  left paren x plus 3 right paren left paren x minus 1 right paren equals 0  in Problem 1. The teacher should draw her attention to the similarity in her approaches to this problem and the previous  A times B equals 0  problem.

Next, ask the student to solve a quadratic equation such as  x squared minus 5 x plus 4 equals 0   The student would likely factor it correctly, as she has already shown her ability to factor. This time, as she set each factor to equal zero, the teacher would have her explain why she is doing that. Next, the student would be given an equation such as  x squared plus 5 x equals negative 6  Drawing the student's attention back to the original  A times B problems,  the teacher would elicit the observation from the student that it was easier to solve the equation when the product was zero rather than another number because she knew that one of the factors would have to equal zero, and then guide the student to write an equivalent expression with a product of zero before solving.

Working from what she already knows (how to factor), the student will begin to build a viable argument for factoring as a solution method for quadratic equations as she comprehends the mathematical concept (the Zero Product Property) that underlies all the problems, from simple to more complex. To help her continue to build a viable argument, the teacher should give her a few more problems (some set up with the product set at zero, some with a product set as non-zero) and have her explain her steps and solutions.

### Performance Characteristics for Constructed-Response Item

The following characteristics guide the scoring of responses to the constructed-response assignment.

 Completeness The degree to which the response addresses all parts of the assignment The degree to which the response demonstrates the relevant knowledge and skills accurately and effectively The degree to which the response provides appropriate examples and details that demonstrate sound reasoning

### Score Scale for Constructed-Response Item

A score will be assigned to the response to the constructed-response item according to the following score scale.

Score Point Score Point Description
4 The "4" response reflects a thorough command of the relevant knowledge and skills:
• The response thoroughly addresses all parts of the assignment.
• The response demonstrates the relevant knowledge and skills with thorough accuracy and effectiveness.
• The response is well supported by relevant examples and details and thoroughly demonstrates sound reasoning.
3 The "3" response reflects a general command of the relevant knowledge and skills:
• The response generally addresses all parts of the assignment.
• The response demonstrates the relevant knowledge and skills with general accuracy and effectiveness.
• The response is generally supported by some examples and/or details and generally demonstrates sound reasoning.
2 The "2" response reflects a partial command of the relevant knowledge and skills:
• The response addresses all parts of the assignment, but most only partially; or some parts are not addressed at all.
• The response demonstrates the relevant knowledge and skills with partial accuracy and effectiveness.
• The response is partially supported by some examples and/or details or demonstrates flawed reasoning.
1 The "1" response reflects little or no command of the relevant knowledge and skills:
• The response minimally addresses the assignment.
• The response demonstrates the relevant knowledge and skills with minimum accuracy and effectiveness.
• The response is minimally supported or demonstrates significantly flawed reasoning.
U The response is unscorable because it is unrelated to the assigned topic or off task, unreadable, written in a language other than English or contains an insufficient amount of original work to score.
B No response.