1. Lesson 2.3 Real Zeros of Polynomial Functions
Essential Question: How do you divide a polynomial by another polynomial?
How do you use polynomial division to find the rational and real zeros of polynomials?

2. Before we start…
Use long division and divide 342 by 2.

3. What is a polynomial function?
Let n be a nonnegative integer and let 𝑎 𝑛 , 𝑎 𝑛−1 , …, 𝑎 2 , 𝑎 1 , 𝑎 0 be real numbers with 𝑎 0 ≠0. The function given by
𝑓 𝑥 = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 +…+ 𝑎 2 𝑥 2 + 𝑎 1 𝑥+ 𝑎 0
Is called a polynomial function of x of degree n

4. Polynomial Long Division
Long division of polynomials looks just like long division of whole numbers.
When you divide a polynomial f(x) by a divisor d(x), you get a quotient polynomial q(x) and a remainder polynomial r(x).

5. Long Division of Polynomials

6. The Division Algorithm
The degree of the remainder must be less than the degree of the divisor.

7. Long Division of Polynomials
The Division Algorithm can also be written as
In the Division Algorithm, the rational expression f (x)/d (x) is improper because the degree of f (x) is greater than or equal to the degree of d (x). On the other hand, the rational expression r (x)/d (x) is proper because the degree of r (x) is less than the degree of d (x).

8. How do you divide a polynomial by another polynomial?
Write polynomial division in the same format you use when dividing numbers.
Include 0 as the coefficient of any “missing” terms in the dividend
At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor.

9. Divide 6𝑥3 – 19𝑥2 + 16𝑥 – 4 by 𝑥 – 2.

10. Divide 2𝑥3 –5𝑥2 + 𝑥 –8 by 𝑥 –3.

11. Divide 𝑥3 −2𝑥2 −9 by 𝑥 −3.

12. Divide 8𝑥3 −1 by 2𝑥−1.

13. Polynomial Synthetic Division
Synthetic division can be used to divide any polynomial by a divisor of the form x – k, so a linear divisor.
When you using synthetic division, you are actually evaluating the polynomial at k.

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

15. To use synthetic division
Write the coefficients of f(x) in order of descending exponents.
Include 0 as the coefficient of any “missing” terms in the dividend
Write the value at which f(x) is being evaluated to the left.
Remember to use the opposite of k.
Bring down the leading coefficient.
Multiply the leading coefficient by the x-value.
Write the product under the second coefficient.
Add.
Then repeat the process for all remaining coefficients.

16. Use synthetic division to divide
𝑥4 − 10𝑥2 − 2𝑥 + 4 by 𝑥 + 3.

17. Use synthetic division to divide
5 𝑥 3 +8 𝑥 2 −𝑥 +6 by 𝑥+2.

18. Use synthetic division to divide
𝑥 2 +8𝑥+1 by 𝑥+4.

19. Use synthetic division to divide
𝑥 4 −5 𝑥 3 −8 𝑥 2 +13𝑥 −12 by 𝑥−6.

20. Remainder Theorem
If a polynomial 𝑓(𝑥) is divided by 𝑥 – 𝑘, then the remainder is 𝑟 = 𝑓(𝑘).
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).

21. Use the Remainder Theorem to evaluate the following function at 𝑥=−2

22. Use the Remainder Theorem to evaluate the following function at 𝑥=−1

23. Use the Remainder Theorem to evaluate the following function at 𝑥=1

24. Factor Theorem
A polynomial 𝑓(𝑥) has a factor 𝑥−𝑘 if and only if 𝑓(𝑘) = 0.
This theorem states that you can test whether a polynomial has (𝑥−𝑘) as a factor by evaluating the polynomial at 𝑥 = 𝑘. If the result is 0, then (𝑥−𝑘) is a factor.

25. The Factor Theorem Problem
The factor theorem can be used to solve a variety of problems.
Problem
Given one factor of a polynomial, find the other factors.
Given one zero of a polynomial function, find the other zeros.
Given one solution of a polynomial equation, find the other solutions.

26. 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).

27. Show that (𝑥+3) are factors of
𝑓 𝑥 = 𝑥 3 −19𝑥−30.
Then find the remaining factors of 𝑓 (𝑥).

28. Show that (𝑥−1) are factors of
𝑓 𝑥 = 𝑥 4 −1.
Then find the remaining factors of 𝑓 (𝑥).

29. The Remainder and Factor Theorems

30. Rational Zero Theorem
The Rational Zero Theorem is a way to find the zeros of any polynomial function, even when a zero or factor is not given.
The Rational Zero Theorem looks at the factors of the leading coefficient and the constant term.

31. The Rational Zero Test

32. Rational Zero Theorem and Finding Zeros
Find the factors of the constant term
Find the factors of the leading coefficient
List possible rational zeros using theorem
Simplify the list of possible zeros

33. Find the rational zeros of 𝑓 (𝑥) = 𝑥3 + 𝑥 + 1

34. Find the rational zeros of 𝑓 𝑥 = 𝑥 3 −7𝑥−6

35. Find the rational zeros of 𝑓 𝑥 =4 𝑥 4 − 𝑥 3 −3 𝑥 2 +9𝑥−10

36. Other Tests for Zeros of Polynomials
You know that an nth-degree polynomial function can have at most n real zeros. Of course, many nth-degree polynomials do not have that many real zeros. For instance, 𝑓 (𝑥) = 𝑥2 + 1 has no real zeros, and 𝑓 (𝑥) = 𝑥3 + 1 has only one real zero. The following theorem, called Descartes’s Rule of Signs, sheds more light on the number of real zeros of a polynomial.

37. Descartes’ Rule of Signs
French mathematician Rene Descartes found the following relationship between the coefficients of a polynomial function and the number of positive and negative zeros of the function.

38. Descartes’ Rule of Signs
Let f (x) be a polynomial function with real coefficients.
The number of positive real zeros of f is equal to the number of changes in sign of the coefficients of f (x) or is less than this by an even integer.
The number of negative real zeros of f is equal to the number of changes in sign of the coefficients of f (-x) or is less than this by an even integer.

39. A variation in sign means that two consecutive (nonzero) coefficients have opposite signs. When using Descartes’s Rule of Signs, a zero of multiplicity k should be counted as zeros. For instance, the polynomial x3 – 3x + 2 has two variations in sign, and so has either two positive or no positive real zeros. Because 𝑥3 – 3𝑥 + 2 = (𝑥 – 1)(𝑥 – 1)(𝑥 + 2) you can see that the two positive real zeros are x = 1 of multiplicity 2.

40. Describe the possible real zeros of 𝑓 𝑥 =3 𝑥 3 −5 𝑥 2 +6𝑥−4

41. Describe the possible real zeros of 𝑓 𝑥 =−2 𝑥 3 +5 𝑥 2 −𝑥+8

42. Describe the possible real zeros of 𝑓 𝑥 =2 𝑥 3 − 𝑥 2 +𝑥+4

43. Other Tests for Zeros of Polynomials
Another test for zeros of a polynomial function is related to the sign pattern in the last row of the synthetic division array.
This test can give you an upper or lower bound of the real zeros of f, which can help you eliminate possible real zeros. A real number c is an upper bound for the real zeros of f when no zeros are greater than c Similarly, c is a lower bound when no real zeros of f are less than c.

44. Upper and Lower Bound Rules
Let 𝑓 𝑥 be a polynomial with real coefficients and a positive leading coefficient. Suppose 𝑓 𝑥 is divided by 𝑥−𝑐, using synthetic division.
1. If 𝑐>0 and each number in the last row is either positive or zero, then c is an upper bound for the real zeros of f.
2. If 𝑐<0 and the numbers in the last row are alternately positive and negative (zero entries count as positive or negative), then c is a lower bound for the real zeros of f.

45. Find all the real zeros of 𝑓 𝑥 =10 𝑥 4 −15 𝑥 3 −16 𝑥 2 +12𝑥

46. Find all the real zeros of 𝑓 𝑥 = 𝑥 4 −6 𝑥 3 +6 𝑥 2 +10𝑥−3

47. Find all the real zeros of 𝑓 𝑥 =3 𝑥 5 −4𝑥 4 −10 𝑥 3 +10 𝑥 2 +7𝑥−6

48. How do you divide a polynomial by another polynomial
How do you divide a polynomial by another polynomial? How do you use polynomial division to find the rational and real zeros of polynomials?

49. Ticket Out the Door
How many possible zeros does this function have? 𝑓 𝑥 =−4 𝑥 7 −3 𝑥 5 + 𝑥 3 −2𝑥