1. Lesson 1: Successive Differences in Polynomials
7 minutes, with explanation (2 slides)
A-SSE.A.2
A-APR.C.4
Lesson 1: Successive Differences in Polynomials
Opening Exercise
John noticed patterns in the arrangement of numbers in the table below:
Number: (𝑥)
2.4
3.4
4.4
5.4
6.4
Square: 𝑓 𝑥 = 𝑥 2
5.76
11.56
19.36
29.16
40.96
First Differences
5.8
7.8
9.8
11.8
Second Differences 2
Module 1 Topic A: Polynomials – From Base 10 to Base X
Pearson Text: Section 5-1
Assuming that the pattern would continue, he used it to find the value of
Explain how he used the pattern to find , and then use the pattern to find

2. Explanation of the Opening Exercise
How did he use the table to find the value of ?
By following the pattern in the first differences, he just added 13.8 to
This results in =54.76, which is true.
How can we continue the pattern to find ?
If the pattern continues to hold, the next first difference is Add and we get that =70.56, which is true.

3. Discussion 3 minutes (1 slide)
Let the sequence 𝑎 0 , 𝑎 1 , 𝑎 2 , 𝑎 3 , … be generated by evaluating a polynomial expression at the values 0, 1, 2, 3,… . The numbers found by evaluating 𝑎 1 − 𝑎 0 , 𝑎 2 − 𝑎 1 , 𝑎 3 − 𝑎 2 , … form a new sequence which we will call the first differences of the polynomial. The differences between successive terms of the first differences are called the second differences (and so on).
Definition Example: Let 𝑓 𝑥 = 𝑥 We can look at a list of values we get for plugging in different values for 𝑥: 𝒙
𝒇 𝒙 = 𝒙 𝟐
Evaluate
1st Diff
2nd Diff
𝑓 0 = 0 2
1−0=𝟏 1
𝑓 1 = 1 2
3−1=𝟐
4−1=𝟑 2
𝑓 2 = 2 2 4
5−3=𝟐
9−4=𝟓 3
𝑓 3 = 3 2 9
Delete the white rectangles to see notes; otherwise the animation is there to reveal the table piece-by-piece.

4. Lesson: Example 1 4 minutes (1 slide)
What is the sequence of first differences for the linear polynomial given by 𝑎𝑥+𝑏, where 𝑎 and 𝑏 are constant coefficients?
Start by identifying: What is the variable that I can replace above? 𝑥
Pick some numbers: What makes sense to replace for 𝑥?
1, 2, 3, …
List the terms: What are the terms of my sequence?
𝑎 1 +𝑏=1a+b 𝑎 2 +𝑏=2𝑎+𝑏 𝑎 3 +𝑏=3𝑎+𝑏 …
Find the 1st differences: What are the terms of my sequence?
2a+b −(a+b) 3𝑎+𝑏 −(2𝑎+𝑏)
List the 1st differences: 𝑎, 𝑎, 𝑎, 𝑎, …
The first differences of a linear expression are constant. This is essentially the slope of a line. You are saying that from one point to the next on a linear function, there is a constant rate of change.

5. Lesson: Example 2 5 minutes (3 slides)
Try these on your own, then move to the next slide for the first differences
Find the first, second, and third differences of the polynomial 𝑎 𝑥 2 +𝑏𝑥+𝑐 by filling in the blanks in the following table. 𝒙
𝒂 𝒙 𝟐 +𝒃𝒙+𝒄
First Differences
Second Differences
Third Differences 𝑐 1
𝑎+𝑏+𝑐 2
4𝑎+2𝑏+𝑐 3
9𝑎+3𝑏+𝑐 4
16𝑎+4𝑏+𝑐 5
25𝑎+5𝑏+𝑐

6. Example 2, Continued. 𝒙
𝒂 𝒙 𝟐 +𝒃𝒙+𝒄
First Differences 𝑐
𝑎+𝑏 1
𝑎+𝑏+𝑐
3𝑎+𝑏 2
4𝑎+2𝑏+𝑐
5𝑎+𝑏 3
9𝑎+3𝑏+𝑐
7𝑎+𝑏 4
16𝑎+4𝑏+𝑐
9𝑎+𝑏 5
25𝑎+5𝑏+𝑐
Remember, to find the difference between terms, you should subtract any given term away from the following term.
𝑡 1 − 𝑡 0
𝑎+𝑏+𝑐 − 𝑐
𝑎+𝑏
The difference between the term where 𝑥=1 ( 𝑡 1 ) and the term where 𝑥=0 ( 𝑡 0 ) is 𝑎+𝑏
Keep the process going to find the remaining first differences.
What do you notice about the type of expression given by the first differences of a quadratic expression?

7. Example 2, finished 𝒙 𝒂 𝒙 𝟐 +𝒃𝒙+𝒄 First Differences Second Differences
Third Differences 𝑐
𝑎+𝑏 1
𝑎+𝑏+𝑐 2𝑎
3𝑎+𝑏 2
4𝑎+2𝑏+𝑐
5𝑎+𝑏 3
9𝑎+3𝑏+𝑐
7𝑎+𝑏 4
16𝑎+4𝑏+𝑐
9𝑎+𝑏 5
25𝑎+5𝑏+𝑐

8. Thinking about example 3
7 minutes, with preparation and explanation (4 slides)
Thinking about example 3
We just spent some time developing the first, second, and third differences of a second degree polynomial.
When our expression was of the form 𝑎 𝑥 2 +𝑏𝑥+𝑐, we could describe the expression and its differences as follows:
The expression itself: quadratic
First differences: linear did you notice how all of the first differences could have been slope-intercept form lines?
Second differences: constant For a second degree polynomial, all the second differences appeared to be 2𝑎
Third differences: zero The third differences of a second degree polynomial zeroed out! Interesting!
BEFORE WE MOVE ON:
Let’s come up with an idea for how we think third degree polynomials will work out. Can you make a conjecture about what the first differences will look like? When do you think the differences in a third degree polynomial become linear?
What do you think is the constant term that they eventually break down into? When will that happen?

9. Lesson: Example 3
Find the second, third, and fourth differences of the polynomial 𝑎 𝑥 3 +𝑏 𝑥 2 +𝑐𝑥+𝑑 by filling in the following table. 𝒙
𝒂 𝒙 𝟑 +𝒃 𝒙 𝟐 +𝒄𝒙+𝒅
First Differences
Second Differences
Third Differences
Fourth Differences 𝑑
𝑎+𝑏+𝑐 1
𝑎+𝑏+𝑐+𝑑
6𝑎+2𝑏
7𝑎+3𝑏+𝑐 6𝑎 2
8𝑎+4𝑏+2𝑐+𝑑
12𝑎+2𝑏
19𝑎+5𝑏+𝑐 3
27𝑎+9𝑏+3𝑐+𝑑
18𝑎+2𝑏
37𝑎+7𝑏+𝑐 4
64𝑎+16𝑏+4𝑐+𝑑
24𝑎+2𝑏
61𝑎+9𝑏+𝑐 5
125𝑎+25𝑏+5𝑐+𝑑

10. Discussion: example 3
From Example 1, the first differences of a first-degree polynomial (linear) were all: 𝑎 From Example 2, the second differences of a second-degree polynomial (quadratic) were all: 2𝑎 From Example 3, the third differences of a third-degree polynomial (cubic) were all: 6𝑎 YOUR TURN! Make a conjecture about the fourth differences of a fourth-degree polynomial (quartic).

11. We read this as “n-factorial times a”.
Discussion: example 3
If you said 24𝑎, you’d be correct! 1 ∙2∙3∙𝟒=24, and our leading term always kept its 𝑎. In fact, the 𝒏 𝒕𝒉 difference of an 𝒏 𝒕𝒉 degree polynomial is going to be 𝒏! 𝒂. A factorial of a positive integer 𝑛, written 𝑛! , is given by the product of all the integers between 1 and 𝑛 inclusive. For example, 5!=5∙4∙3∙2∙1 There are a few additional quirks with factorials, but we will cover those at a later time.
The ! in that phrase is called a “bang” and stands for the factorial of an expression.
We read this as “n-factorial times a”.

12. Lesson: Example 4 – Generating polynomials
7 minutes, with explanation (5 slides)
Lesson: Example 4 – Generating polynomials
Very often, we will be presented with data, but not told the model that it fits. When you were in Algebra I, you studied conditions for establishing a linear or exponential relationship. Either the first differences were constant (linear) or the “first factors” were constant (exponential). Now we have a way of establishing higher degree ( 𝑛 𝑡ℎ degree) polynomials by the fact that their 𝑛 𝑡ℎ difference is contant.

13. Lesson: Example 4
What type of relationship does the set of ordered pairs (𝒙, 𝒚) satisfy? How do you know? Fill in the blanks in the table below to help you decide. (The first differences have already been computed for you) 𝒙 𝒚
First Differences
Second Differences
Third Differences 2 −1 1 5 6 17 3 23 35 4 58 59
117

14. Example 4, finish the table𝒙 𝒚
First Differences
Second Differences
Third Differences 2 −1 1 6 5 12 17 3 23 18 35 4 58 24 59
117
Since our third differences are constant, this must be a third degree polynomial (cubic): 𝒂 𝒙 𝟑 +𝒃 𝒙 𝟐 +𝒄𝒙+𝒅

15. Example 4, establishing the polynomial
This last piece is the most important: we now have to use what we know about 𝑛 𝑡ℎ differences to generate the polynomial from the information we have about the points it uses. We have established the following: Our expression is a cubic of the form 𝑎 𝑥 3 +𝑏 𝑥 2 +𝑐𝑥+𝑑 The constant expression of a cubic is 6𝑎 Since the constant of this polynomial is 6, we can say 6𝑎=6, and solve. 𝑎=1 Also, since (0, 2) is a point that satisfies this polynomial, plugging 0 in for 𝑥 lets all the 𝑥 terms drop out, and we get that 𝑑=2. Our polynomial function can now be expressed by: 𝑦= 𝑥 3 +𝑏 𝑥 2 +𝑐𝑥+2 By using other coordinate pairs (points of the form (𝑥,𝑦) ) from the table, we can solve a system of equations to establish our polynomial function.
At this point, students should attempt to use this information to solve for the polynomial expression.

16. Solving Systems of Equations
We are going to substitute two sets of points into our new equation: (1,1) and (2,6) 1 = 1 3 +𝑏 1 2 +𝑐 =1+𝑏+𝑐+2 6 = 2 3 +𝑏 2 2 +𝑐 2 +𝑑 6=8+4𝑏+2𝑐+2 If we subtract two times the first equation from the second, we get: 6=8+4𝑏+2𝑐+2 − (2=2+2b+2c+4) 4=6+2𝑏+0 −2 Solving this, we can see 𝑏=0. Substitue 𝑏=0 back into the first equation to get 1=1+0+𝑐+2, and we can solve so 𝑐=−2. We now have all of our coefficients! 𝑎=1 ;𝑏=0;𝑐=−2;𝑑=2 Which allows us to create the function containing the given points: 𝑦= 𝑥 3 −2𝑥+2

17. 7 minutes
Closing
Key ideas from today: Sequences whose second differences are constant satisfy a quadratic relationship. Sequences whose third differences are constant satisfy a cubic relationship. We can use a combination of 𝑛 𝑡ℎ differences and known points to generate a polynomial function.
Please work on the Exit Slip for the remainder of the period. Complete in full.

18. Problem Set
Page 1 of 2

19. Problem Set (page 2 of 2)