1. Polynomial and Synthetic Division
Pre-Calculus

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

3. Polynomial Division
Subtract!!
Subtract!!
Subtract!!

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

5. Polynomial Division

6. Try This
Example:
Answer:

7. What does it mean if a number divides evenly into another??

8. 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.

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

10. 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

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

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

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

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

15. 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

16. 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.

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

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

19. 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.

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

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

22. 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).

23. 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

24. 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.

25. 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).

26. Long Division of Polynomials

27. The Remainder and Factor Theorems

28. The Remainder and Factor Theorems
The remainder obtained in the synthetic division process has an important interpretation, as described in the Remainder Theorem.
The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. That is, to evaluate a polynomial function f (x) when x = k, divide f (x) by x – k. The remainder will be f (k).

29. Example Using the Remainder Theorem
Use the Remainder Theorem to evaluate the following function at x = –2.
f (x) = 3x3 + 8x2 + 5x – 7
Solution:
Using synthetic division, you obtain the following.

30. Example – Solution
cont’d
Because the remainder is r = –9, you can conclude that
f (–2) = –9.
This means that (–2, –9) is a point on the graph of f. You can check this by substituting x = –2 in the original function.
Check:
f (–2) = 3(–2)3 + 8(–2)2 + 5(–2) – 7
= 3(–8) + 8(4) – 10 – 7
= –9
r = f (k)

31. The Remainder and Factor Theorems
Another important theorem is the Factor Theorem, stated below.
This theorem states that you can test to see whether a polynomial has (x – k) as a factor by evaluating the polynomial at x = k.
If the result is 0, (x – k) is a factor.

32. 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.

33. 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.

34. The Remainder and Factor Theorems
For instance, if you find that x – k divides evenly into f (x) (with no remainder), try sketching the graph of f.
You should find that (k, 0) is an x-intercept of the graph.

35. Can you… 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.

36. Questions?
OK, what if you are given a polynomial to factor without any clues as to possible roots?
Whatcha gonna do?

37. Rational Root Theorem …

38. Homework
The Usual for section 2-3