1. Polynomial Long Division and Synthetic Division

2. What You Should Learn
Use long division to divide polynomials by other polynomials.
Use synthetic division to divide polynomials by binomials of the form (x – k).
Use the Remainder Theorem and the Factor Theorem.

3. When our factoring techniques do not easily work…
Analyzing and Graphing a Function
Let’s say we want to analyze this function and graph it:
f(x) = x7 - 8x5 - 2x4 - 21x3 + 10x x + 72
We know the left and right behavior
We know the y- intercept
To get a good approximation of the graph, we need to know the x-intercepts or the “zeros”. To find all the real zeros of the function we must factor it completely.
Determining if one polynomial is a factor of another polynomial
Factoring a polynomial
Polynomial division will help with this.

4. Today We are going to learn about the process of division
Learn about a couple of theorems to help in factoring and solving higher level polynomials

5. Division of Polynomials
In this section, we will study two procedures for dividing polynomials.
These procedures are especially valuable in factoring and finding the zeros of polynomial functions.

6. Polynomial Division
Polynomial Division is very similar to long division.
Example:

7. Polynomial Division
Subtract!!
Subtract!!
Subtract!!

8. Polynomial Division Example:
Notice that there is no x term. However, we need to include it when we divide.

9. Polynomial Division

10. Try This
Example:
Answer:

11. Now let’s look at another method to divide…
Why???
Sometimes it is easier…

12. Synthetic Division
Synthetic Division is a ‘shortcut’ for polynomial division that only works when dividing by a linear factor (x + b).
It involves the coefficients of the dividend, and the zero of the divisor.

13. Example
Divide:
Step 1:
Write the coefficients of the dividend in a upside-down division symbol. 1 5 6

14. Example
Step 2:
Take the zero of the divisor, and write it on the left.
The divisor is x – 1, so the zero is 1. 1 1 5 6

15. Example
Step 3:
Carry down the first coefficient. 1 1 5 6 1

16. Example
Step 4:
Multiply the zero by this number. Write the product under the next coefficient. 1 1 5 6 1 1

17. Example
Step 5:
Add. 1 1 5 6 1 1 6

18. Example
Step etc.:
Repeat as necessary 1 1 5 6 1 6 1 6 12

19. Example
The numbers at the bottom represent the coefficients of the answer. The new polynomial will be one degree less than the original. 1 1 5 6 1 6 1 6 12

20. Synthetic Division
The pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higher-degree polynomials is similar.)

21. Synthetic Division
This algorithm for synthetic division works only for divisors of the form x – k.
Remember that x + k = x – (–k).

22. Using Synthetic Division
Use synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3.
Solution:
You should set up the array as follows. Note that a zero is included for the missing x3-term in the dividend.

23. Example – Solution
cont’d
Then, use the synthetic division pattern by adding terms in columns and multiplying the results by –3.
So, you have .

24. Try These Examples: Answers: (x4 + x3 – 11x2 – 5x + 30)  (x – 2)
(x4 – 1)  (x + 1) [Don’t forget to include the missing terms!]
Answers:
x3 + 3x2 – 5x – 15
x3 – x2 + x – 1

25. Application of Long Division
To begin, suppose you are given the graph of
f (x) = 6x3 – 19x2 + 16x – 4.

26. Long Division of Polynomials
Notice that a zero of f occurs at x = 2.
Because x = 2 is a zero of f, you know that (x – 2) is a factor of f (x). This means that there exists a second-degree polynomial q (x) such that
f (x) = (x – 2)  q(x).
To find q(x), you can use long division.

27. Example - Long Division of Polynomials
Divide 6x3 – 19x2 + 16x – 4 by x – 2, and use the result to factor the polynomial completely.

28. Example 1 – Solution Think Think Think Multiply: 6x2(x – 2). Subtract.

29. Example – Solution From this division, you can conclude that
cont’d
From this division, you can conclude that
6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2)
and by factoring the quadratic 6x2 – 7x + 2, you have
6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2).

30. Example 1 – Solution
cont’d
Note that this factorization agrees with the graph shown in Figure 2.28 in that the three x-intercepts occur at x = 2, x = , and x = .
Figure 2.28

31. Long Division of Polynomials
In Example 1, x – 2 is a factor of the polynomial 6x3 – 19x2 + 16x – 4, and the long division process produces a remainder of zero. Often, long division will produce a nonzero remainder.
For instance, if you divide x2 + 3x + 5 by x + 1, you obtain the following.

32. Long Division of Polynomials
In fractional form, you can write this result as follows.
This implies that
x2 + 3x + 5 = (x + 1)(x + 2) + 3
which illustrates the following theorem, called the Division Algorithm.
Multiply each side by (x + 1).

33. Long Division of Polynomials

34. Example – Factoring a Polynomial: Repeated Division
Show that (x – 2) and (x + 3) are factors of
f (x) = 2x4 + 7x3 – 4x2 – 27x – 18.
Then find the remaining factors of f (x).
Solution: Using synthetic division with the factor (x – 2), you obtain the following.
0 remainder, so f (2) = 0
and (x – 2) is a factor.

35. Example – Solution
cont’d
Take the result of this division and perform synthetic division again using the factor (x + 3).
Because the resulting quadratic expression factors as
2x2 + 5x + 3 = (2x + 3)(x + 1)
the complete factorization of f (x) is
f (x) = (x – 2)(x + 3)(2x + 3)(x + 1).
0 remainder, so f (–3) = 0
and (x + 3) is a factor.