1. 2.3: Polynomial Division Objectives:
To divide polynomials using long and synthetic division
To apply the Factor and Remainder Theorems to find real zeros of polynomial functions

2. Vocabulary
As a class, use your vast mathematical knowledge to define each of these words without the aid of your textbook.
Quotient
Remainder
Dividend
Divisor
Divides Evenly
Factor

3. Exercise 1
Use long division to divide 5 into 3462. - - -

4. Exercise 1 Use long division to divide 5 into 3462. Quotient Divisor
Dividend - - -
Remainder

5. Exercise 1 Use long division to divide 5 into 3462. Dividend Remainder
Divisor
Divisor
Quotient

6. Remainders
If you are lucky enough to get a remainder of zero when dividing, then the divisor divides evenly into the dividend. This means that the divisor is a factor of the dividend. For example, when dividing 3 into 192, the remainder is 0. Therefore, 3 is a factor of 192.

7. Dividing Polynomials
Dividing polynomials works just like long division. In fact, it is called long division!
Before you start dividing:
Make sure the divisor and dividend are in standard form (highest to lowest powers).
If your polynomial is missing a term, add it in with a coefficient of 0 as a place holder.

8. How many times does x go into x2?
Exercise 2
Divide x + 1 into x2 + 3x + 5 Line up the first term of the quotient with the term of the dividend with the same degree.
How many times does x go into x2?
Multiply x by x + 1 - - - -
Multiply 2 by x + 1

9. Exercise 2 Divide x + 1 into x2 + 3x + 5 Quotient Dividend Divisor- - - -
Divisor
Remainder

10. Exercise 2 Divide x + 1 into x2 + 3x + 5 Dividend Remainder Divisor
Quotient
Divisor

11. Exercise 3
Divide 6x3 – 16x2 + 17x – 6 by 3x – 2

12. Exercise 4
Use long division to divide x4 – 10x2 + 2x + 3 by x – 3

13. Synthetic Division
When your divisor is of the form x - k, where k is a constant, then you can perform the division quicker and easier using just the coefficients of the dividend. This is called fake division. I mean, synthetic division.

14. Synthetic Division Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following pattern. k a b c d
= Add terms ka
= Multiply by k a
Remainder
Coefficients of Quotient (in decreasing order)

15. Synthetic Division Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following pattern.
Important Note: You are always adding columns using synthetic division, whereas you subtracted columns in long division. k a b c d
= Add terms ka
= Multiply by k a

16. Synthetic Division Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following pattern.
Important Note: k can be positive or negative. If you divide by x + 2, then k = -2 because x + 2 = x – (-2). k a b c d
= Add terms ka
= Multiply by k a

17. Synthetic Division Synthetic Division (of a Cubic Polynomial)
To divide ax3 + bx2 + cx + d by x – k, use the following pattern.
Important Note: Add a coefficient of zero for any missing terms! k a b c d
= Add terms ka
= Multiply by k a

18. Exercise 5
Use synthetic division to divide x4 – 10x2 + 2x + 3 by x – 3

19. Exercise 6
Evaluate f (3) for f (x) = x4 – 10x2 + 2x + 3.

20. Remainder Theorem
If a polynomial f (x) is divided by x – k, the remainder is r = f (k). This means that you could use synthetic division to evaluate f (5) or f (-2). Your answer will be the remainder.

21. Exercise 7
Divide 2x3 + 9x2 + 4x + 5 by x + 3 using synthetic division.

22. Exercise 8
Use synthetic division to divide f(x) = 2x3 – 11x2 + 3x + 36 by x – 3. Since the remainder is zero when dividing f(x) by x – 3, we can write: This means that x – 3 is a factor of f(x).

23. Factor Theorem
A polynomial f(x) has a factor x – k if and only if f(k) = 0. This theorem can be used to help factor/solve a polynomial function if you already know one of the factors.

24. Exercise 9
Factor f(x) = 2x3 – 11x2 + 3x + 36 given that x – 3 is one factor of f(x). Then find the zeros of f(x).

25. Exercise 10
Given that x – 4 is a factor of x3 – 6x2 + 5x + 12, rewrite x3 – 6x2 + 5x + 12 as a product of two polynomials.

26. Exercise 11
Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42 given that f(7) = 0.

27. Rational Zero Test: we use this to find the rational zeros
for a polynomial f(x). It says that if f(x) is a polynomial of the form:
Then the rational zeros of f(x) will be of the form:
Where p = factor of the constant &
q = factor of leading coefficient
Rational zero =
Possible rational zeros = factors of the constant term___
factors of the leading coefficient
Keep in mind that a polynomial can have rational zeros, irrational zeros and complex zeros.

28. Ex 1: Find all of the possible rational zeros of f(x)

29. Ex 2: Find the rational zeros of:
Let’s start by listing all of the possible rational zeros, then we will use synthetic division to test out the zeros:
1. Start with a list of factors of -6 (the constant term): p =
2. Next create a list of factors of 1 (leading coefficient): q =
3. Now list your possible rational zeros: p/q =
Testing all of those possibilities could take a while so let’s use the graph of f(x) to locate good possibilities for zeros.
Use your trace button!

30. Ex 2 continued: Find all of the rational zeros of the function

31. Ex 3: Find all the real zeros of :
p = Factors of 3:
q = Factors of 2:
Candidates for rational zeros: p/q =
Let’s look at the graph: Which looks worth trying?
Now use synthetic division to test them out.

32. Homework
Dividing Polynomials Worksheet
Page
36,38, odd