Unit 2 Test                                

Name: _________________ Teacher: Tyrnee Bevell
Date: 10/14/2021 Class: Secondary 3 Honors
Description: Test on Solving Polynomials.
5 4 3 2 1
1 Listed all p/q correctly: (plus or minus) 6/5, 6, 3/5, 3, 2/5, 2, 1/5, 1 Listed most of the p/q options but forgot a couple. Answer was not p/q but they had the possible combinations of positive, negative, and imaginary roots.
2 Solved the polynomial correctly and got that x=-3 (given) and found the other two zeros: x=1, 2/5 Recognized that -3 is a root and used it to do synthetic division. Solved the remaining quadratic incorrectly. Recognized that -3 is a root and used it to do synthetic division.
3 Has x-intercepts at -3, 2/5, and 1. End behavior is up on the right side and down on the left side. The shape of the graph is correct and crosses the x-axis at one of the zeros correctly. The shape of the graph is correct but doesn't cross the x-axis at the correct spots.
4 The student correctly identifies 1-root2 as the other root. The student correctly identifies 1-root2 as the other root but also lists -7/9 as a possible root.
5 The student distributes correctly: (x+4)(x^2-10x+26) and gets x^3-6x^2-14x+104 The student writes out the factors (x+4)(x-5-i)(x-5+i), and recognizes the imaginary roots are a sum of squares. Or the student distributes correctly/with few errors without using a sum of squares. The student correctly identifies 5-i as a conjugate root, or forgets about the conjugate and distributes correctly with only the two roots.
6 The student correctly writes the equation: -(1/6)(x+5)^3 The student incorrectly writes the equation: (-1/6x+5)^3 or -(1/6)x^3-5 The student has two of the 4 features correct.
7 The student correctly identifies all 4 zeros at -4, 1, 3, 3/2. The student correctly identifies all 3 of the 4 zeros at -4, 1, 3, 3/2. The student correctly identifies all 2 of the 4 zeros at -4, 1, 3, 3/2. The student correctly identifies all 1 of the 4 zeros at -4, 1, 3, 3/2.
8 The student gets the two roots to be + or - 5i/4. The student gets the two roots to be + or - 5/4. The student factors in either the form of sum (correct) or difference of squares, or solves the equation by subtracting 25 over to the 0, dividing by 16, taking the square root, and getting the answer of 5/4 but forgetting the plus or minus sign.
9 The student uses a graphing calculator to find the zero at -8.7. The student has the answer 11 or -11 (just to be kind).
10 The student correctly selects c as the answer. The student selects a as the answer. The student shows some work off to the side even though it leads to the wrong answer.
11 The student correctly selects b as the answer. The student selects c as the answer. The student shows some work off to the side even though it leads to the wrong answer.
12 The student correctly selects f as the answer. The student selects b, c, e as the answer. The shape is right but the transformations are wrong. The student selects d. The shape is wrong but the transformations are close to being correct (right and down).
13 The explanation references the quadratic formula or square roots having two answers: the + or - version of each. The students explains that there always needs to be two answers for imaginary and irrational answers but doesn't explain how or why. The student has an explanation written.
14 The student identifies zeros at 0 (mult.2), -4, -2(mult.4). Graphs the function up on the left, down through -4, up to -2 and bounce down, up to 0 and bounce down to being down on the right side. The end behavior is as x neg inf, y pos inf, and x pos inf, y neg inf. The student identifies the zeros without multiplicity, or graphs the function without bounces, or The end behavior is as x neg inf, y neg inf, and x pos inf, y pos inf. The student does 2 of the following: The student identifies the zeros without multiplicity, or graphs the function without bounces, or The end behavior is as x neg inf, y neg inf, and x pos inf, y pos inf. The student does 3 of the following: The student identifies the zeros without multiplicity, or graphs the function without bounces, or The end behavior is as x neg inf, y neg inf, and x pos inf, y pos inf. There are few answers or only one part of the question is done.