
HW for 11/13
Posted by Allen Olsen on 11/13/2018Objectives: Solve rational equations in one variable.
HW: Sect. 2.7, pp. 227  30:
1  29 (odd), 51.

No HW for 11/5
Posted by Allen Olsen on 11/5/2018Today you took a formative assessment of your understanding of graphing rational functions.
Per district policy: No homework tonight.

HW for 11/1 (A), 11/2 (D)
Posted by Allen Olsen on 11/1/2018Objective: Use graphical transforms and polynomial division to graph rational functions.
Read pages 216  216 in Sect. 2.6, especially examples 1  5. Then, on pages 220  21, do
Problems 11  18, 19  29 (odd), 31  36.

HW for 10/31
Posted by Allen Olsen on 10/31/2018Objective: Understand the logical process for factoring polynomial functions using synthetic division.
HW: p. 220: Quick Review 2.6, problems 1  10; Section 2.6 Exercises, problems 1  10.

HW for 10/30
Posted by Allen Olsen on 10/30/2018Objective: Find complex zeros and factors of polynomials.
HW: Sect. 2.5, pp. 209  10:
17  20; 33  42; 28, 30  32.

HW for 10/29
Posted by Allen Olsen on 10/29/2018Objective: Add, multiply and divide complex numbers. Find complex zeros of polynomial functions.
HW: Sect. 2.5, pp. 209  10:
1  11 (odd), 13  16, 27, 29.

HW for 10/25 (A), 10/26 (D)
Posted by Allen Olsen on 10/25/2018You took a quiz today on Sect. 2.4, the use of Polynomial Division to factor polynomials.
HW: Read Sect. P.6 on complex numbers, how to add them, how to multiply them, how to divide them, and how to find complex solutions of a quadratic equation using the quadratic formula.
On pages 51  52, do problems 1  7 odd, 9  16, 17, 19, 33  39 (odd).

HW for 10/24
Posted by Allen Olsen on 10/24/2018Objective: Prepare for the next Quiz in the next class period: Thursday (A) or Friday (D).
Topics: Long Polynomial Division; Synthetic Division; generating possible integer zeros; Finding upper and lower bounds for zeros; generating possible rational (fractional) zeros; using the zeros you found to factor the polynomial.
Vocab for use in your answers: Factor Theorem, Remainder Theorem; integer, rational, irrational zeros; factor.

HW for 10/23
Posted by Allen Olsen on 10/23/2018Objective: Find factors of polynomials by using synthetic division in the steps detailed below. I have summarized the notes from class at the bottom of this posting.
HW: Sect. 2.4, pp. 200  01:
37  44 (look at example 6); 33  36 (look at example 5).
Quiz Reminder: this Thursday (A) and this Friday (D). It is on 2.1 and 2.4, covering polynomial division, synthetic division, Factor / Remainder theorems, and finding zeros / factoring polynomials.
Notes from Class:
1. "Find a zero, find a factor" by the Factor Theorem. Thus we will be looking for zeros of the polynomial.
2. By the Remainder Theorem, we can test if "a" is a zero of f(x) by dividing f(x) by (xa), using synthetic division.
3. First we look for integer zeros. The only possible integer zeros are factors of the constant term.
4. Make a list of the possible integer zeros, then test them in the order +1, 1, etc., working away from the origin. If at any point the result of the division has all positive signs (+++++), then the value of "a" you are testing is an upper bound, and you don't have to test any higher values. If at any point the result of the division has alternating signs (++++), then the value of "a" that you are testing is a lower bound, and you don't have to test any lower values.
5. If the remainder of the division is zero, then the "a" value is a zero of the function. Continue testing with the quotient, which has a lower degree than the original function.
6. When you finish testing for integer zeros, plot the results of your divisions to make a crude sketch of the function. You are looking for the function to change sign somewhere. When the function changes sign, there is a zero between the two corresponding integers.
7. Now it's time to look for rational zeros, that is, zeros that are fractions. First we make a list of the possible rational zeros. The possible numerators are the factors of the constant term, as before. The possible denominators are the factors of the leading coefficient. First make a list of all of the combinations. There's usually a lot of them, but don't worry, because we will sort them out. You only rarely have to do all of them.
8. First, any fractions that are actually integers, e.g., 8/2 = 4, you don't have to test because we've already done all the integers.
9. Second, if the value of the fractions is outside of the lower and upper bounds that you found in step 4, you don't have to test those fractions.
10. Third, if you found sign changes in step 6 above, try the fractions that fit into that interval.
11. Synthetic division will work with fractions. If you get a zero remainder, then the "a" value is a zero of the function. Continue testing with the quotient. If you didn't get a zero, look for the patterns like in step 4 that indicate an upper or lower bound. You don't have to test beyond these.
12. At this point you have used up the fractions and integers. If you still have more zeros to find, they must be irrational. Hopefully at this point your quotient is down to a quadratic, because you can use the quadratic formula to find irrational zeros.
13. Update your sketch of the function that you started in step 6, recording the exact values of the zeros, and sketching what the function looks like based on your investigation.
Example from class: f(x) = 2x^47x^38x^2+14x+8.
Step 1: The constant term is "8". Its factors are ±1, ±2, ±4, and ±8. First we use synthetic division to find integer zeros.
1 2 7 8 14 8 1 2 7 8 14 8 2 2 7 8 14 8
2 5 13 1 2 9 1 13 4 6 28 28
2 5 13 1 9 2 9 1 13 5 2 3 14 14 20
2 2 7 8 14 8 4 2 7 8 14 8
3 20 24 20 8 4 16 8
2 10 12 10 12 2 1 4 2 0
Step 2: Now we do the analysis suggested by steps 1  6 above. We learn the following:
a. 4 is a zero, and therefore (x4) is a factor: 2x^47x^38x^2+14x+8 = (x4)(2x^3+x^24x2) + 0.
b. 2 is a lower bound, so we don't have to test 4 and 8 from our original list. 8 from our original list turns out to be an upper bound.
c. In terms of sign reversal, we note that there are zeros between 0 and 1; and between 1 and 2. The next step is to look for some rational zeros that might be in these intervals.
Step 3: To construct possible rational zeros, we look at the possible denominators, which are the factors of the leading coefficient, which is 2. The only factors are 1 and 2, but 1 as a denominator doesn't lead to any fractions. 2 leads to the fractions ±1/2, since 2/2, 4/2 and 8/2 are all integers and we are done with integers. We can use synthetic division to test the two fractions. Use the quotient from above: 2x^3+x^24x2.
1/2 2 1 4 2 1/2 2 1 4 2
1 1 3/2 1 0 2
2 2 3 7/2 2 0 4 0
Step 4: Analysis: 1/2 is a zero, so (x + 1/2) is a factor with quotient 2x^2  4. All other zeros must be irrational, and we can use this quotient to find them.
Step 5: Irrational Zeros. 2x^2  4 = 0 has solutions x = ±√(2) = ±1.414. These lie between 1 and 2, and 1 and 2.
Step 6: Now you can draw the function, based on the zeros that we found and the remainders in the synthetic divisions.
Step 7: The factorization is 2x^47x^38x^2+14x+8 = (x4)(2x+1)(x√2)(x+√2).

HW for 10/22
Posted by Allen Olsen on 10/22/2018Objective: Prove the Remainder and Factor Theorems.
HW: Sect. 2.4, pp. 200  01:
Problems 49  56.