1. Unit 4 Polynomials

2. Polynomial Graphs: Domain, Range, Zeros and Extrema

3. Vocabulary Domain: the x-values of the function
Range: they y-values of the function
Zeros: where the graph crosses the x- axis
Extrema:
Relative Maximum – the highest point in a particular section of the graph (like a hill)
Relative Minimum - the lowest point in a particular section of the graph (like a valley)
Vocabulary

4. f(x)=x2 + 2x Domain: (-∞, ∞) Range: [-1, ∞) Zeros: Relative Minimum:
Relative Maximum:
Domain:
(-∞, ∞)
Range:
[-1, ∞)
{-2, 0}
(-1,-1)
none

5. g(x) = -2x2 + x Zeros: Relative Minimum: Relative Maximum: {0, ½} none
Domain:
(-∞, ∞)
Range:
(-∞, ¼]
{0, ½}
none
(¼, ¼)

6. h(x)= x3 - x Domain: (-∞, ∞) Range: Zeros: Relative Minimum:
Relative Maximum:
Domain:
(-∞, ∞)
Range:
{-1, 0, 1}
(½ ,- ½)
(-½ , ½)

7. j(x) = -x3 +2x2 +3x Domain: (-∞, ∞) Range: Zeros: Relative Minimum:
Relative Maximum:
Domain:
(-∞, ∞)
Range:
{-1, 0, 3}
(-½ ,- 1)
(2 , 6)

8. k(x)= x4 -5x2 + 4 Domain: (-∞, ∞) Range: [-2, ∞) Zeros:
Relative Minimum:
Relative Maximum:
Domain:
(-∞, ∞)
Range:
[-2, ∞)
{-2, -1, 1, 2}
(-1½ ,-2), (1½ ,-2)
(0 , 4)

9. l(x) = -(x4 -5x2 + 4) Domain: (-∞, ∞) Range: (-∞, 2] Zeros:
Relative Minimum:
Relative Maximum:
Domain:
(-∞, ∞)
Range:
(-∞, 2]
{-2, -1, 1, 2}
(0,-4)
(-1½ , 2), (1½ , 2)

10. Intervals: Vocabulary
Increase: Where on the graph is the function increasing (going up)
**Use interval notation**
Decrease: Where on the graph is the function decreasing (going down)
Vocabulary

11. Symmetry:
A function is considered EVEN if it has symmetry about the y-axis.
Only even degree polynomials with ALL EVEN exponents can have even symmetry.
Ex: 𝑥 4 +3 𝑥 2 +6
Non-Ex: 𝑥 6 −3𝑥
A function is considered ODD is it has symmetry about the origin.
Only odd degree polynomials with ALL ODD exponents can have odd symmetry.
Ex: 𝑥 5 +3 𝑥 3 +𝑥
Non-Ex: 𝑥 3 −3𝑥+2
Vocabulary

12. f(x)=x2 + 2x Odd/Even/Neither Interval of increase:
Interval of decrease:
Odd/Even/Neither
[-1,∞)
( -∞, -1]

13. g(x) = -2x2 + x Odd/Even/Neither Interval of increase:
Interval of decrease:
Odd/Even/Neither
(-∞, ¼]
[¼, ∞)

14. h(x)= x3 - x Intervals of increase: Interval of decrease:
Odd/Even/Neither
(- ∞, -½]
, [½ , ∞)
[- ½ , ½ ]

15. j(x) = -x3 +2x2 +3x Odd/Even/Neither Intervals of increase:
Interval of decrease:
Odd/Even/Neither
[- ½ , 2]
(- ∞, -½]
, [ 2 ,∞)

16. k(x)= x4 -5x2 + 4 Odd/Even/Neither Intervals of increase:
Interval of decrease:
Odd/Even/Neither
[- 1½ , 0]
, [ 1½, ∞)
(- ∞, -1½]
, [ 0 ,1½)

17. l(x) = -(x4 -5x2 + 4) Intervals of increase: Interval of decrease:
Odd/Even/Neither
(- ∞, -1½]
, [ 0 ,1½)
[- 1½ , 0]
, [ 1½, ∞)

18. Write the standard form of the equation for the polynomial function with the given zeroes
1, 3(multiplicity 2)
f(x) = (x-1)(x-3)(x-3)
f(x) = (x2 -3x – x +3)(x-3)
f(x) = (x2 - 4x + 3)(x–3)
f(x) = x3 - 3x2 - 4x2 + 12x +3x– 9
f(x) = x3 - 7x2+15x-9
Multiplicity means that a zero is used as a factor more than once

19. Long Division

20. Algebraic long division
Divide 2x³ + 3x² - x + 1 by x + 2
x + 2 is the divisor
2x³ + 3x² - x + 1 is the dividend
The quotient will be here.
Algebraic long division

21. Algebraic long division
First divide the first term of the dividend, 2x³, by x (the first term of the divisor).
This gives 2x². This will be the first term of the quotient.
Algebraic long division

22. Algebraic long division
Now multiply 2x² by x + 2
and subtract
Algebraic long division

23. Algebraic long division
Bring down the next term, -x.
Algebraic long division

24. Algebraic long division
Now divide –x², the first term of –x² - x, by x, the first term of the divisor
which gives –x.
Algebraic long division

25. Algebraic long division
Multiply –x by x + 2
Algebraic long division
and subtract

26. Algebraic long division
Bring down the next term, 1
Algebraic long division

27. Algebraic long division
Divide x, the first term of x + 1, by x, the first term of the divisor
Algebraic long division
which gives 1

28. Algebraic long division
Multiply x + 2 by 1
and subtract

29. Algebraic long division
The quotient is 2x² - x + 1
The remainder is –1.
Algebraic long division

30. Algebraic long division
Write the quotient as:

31. Synthetic Division

32. Synthetic Division Another way to divide certain polynomials
The divisor must be a linear (exponent=1) binomial with leading coefficient 1
For example, x-5
Synthetic Division

33. Synthetic Division Example:
Think about how you would set up the long division problem
Make sure that the dividend is in descending order and that no terms are missing
You may have to include some 0 coefficients if terms are missing
Synthetic Division

34. Remove the variables from the dividend leaving the coefficients
Don’t forget any 1’s or 0’s that may be there
Set up the multiplier
Use the divisor and solve it for zero -2
Synthetic Division

35. Synthetic Division -2 -4 2 -2 -1 1 -1
Leave a blank line below the coefficients, and underline it.
Bring down the first coefficient
Use the multiplier, and place the product below the next coefficient.
Add
Repeat 2 -2 -4 2 -2 -1 1 -1
Synthetic Division

36. Synthetic Division -2 -4 2 -2 -1 1 -1
Now, use the bottom line to write the quotient as a polynomial
Place the variables back into the problem
The first coefficient has degree that is one less than in the dividend
The last number is the remainder 2 -2 -4 2 -2 -1 1 -1
Synthetic Division

37. -1 -2 5 -9 2 -5 9
-10
Example 2:

38. 4 4 16 64
244 1 4 16 61
249
Example 3:

39. 1 1 1 1 1 1 1 1 1 1 1
Example 4:

40. Rational Root Theorem

41. To find the zeroes of a Polynomial:
The Rational Root Theorem:
If f(x) = anxn a1x + a0 has integer coefficients, then every rational zero of f(x) has the following form: p =
factor of constant term a0 q
factor of leading coefficient an
To find the zeroes of a Polynomial:

42. What is the constant?
-10
This is going to be the numerator
What is the leading coefficient? 2
This is going to be the denominator
Possible roots:
Use the rational root theorem to determine all possible rational roots:

43. The degree of the polynomial will tell you the max number of zeroes
Some roots may be imaginary
Use factors of the constant term (a0) as your numerator
Use factors of the leading coefficient (an) as your denominator
Use your calculator and synthetic division to determine if the possible ratios are actual zeroes
Things to consider:

44. Find the rational zeroes of:
Possible:
Use your calculator to graph the function and determine which possible zeroes to try.
Which ones look like they may work?
Use synthetic division to check them
Find the rational zeroes of:

45. Actual Solutions 2 -3/2 1/4 Try 2 2 8 -6 -23 6
now use, the smaller polynomial:
Try -3/2
-3/
-12 3

46. Now, you can give the factored form of the equation
The zeroes for
are { 2, -3/2, ¼ }
So, the factored form of the equation is:
Now, you can give the factored form of the equation

47. Find the rational zeroes of:
Possible:
Use your calculator to graph the function and determine which possible zeroes to try.
Which ones look like they may work?
Use synthetic division to check them
Find the rational zeroes of:

48. Actual Solutions 3 -3/2 1/3 Try 3 3 6 -11 -24 9
now use, the smaller polynomial:
Try -3/2
-3/
-9 3

49. Now, you can give the factored form of the equation
The zeroes for
are { 3, -3/2, 1/3 }
So, the factored form of the equation is:
Now, you can give the factored form of the equation

50. Solving Polynomial Equations

51. The degree of the polynomial determines the number of zeros/roots/solutions the polynomial has.
These zeros may be real or complex.
Quadratic Equation:
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Things to Remember…

52. Steps
Example
Use the Rational Root Theorem to find all the rational roots of the polynomial.
𝑥 4 −14 𝑥 2 +45=0 Possible Roots: ±1, ±3, ±5, ±9,±15, ±45 Try: 3, −3
Example 1

53. Steps
Example
Use the remaining polynomial and the quadratic formula to find the last two roots.
𝑥 2 −5=0 𝑎=1 𝑏=0 𝑐=−5 𝑥= −0± 0 2 −4(1)(−5) 2(1) 𝑥= ± 20 2 = ±2 5 2 =± 5
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Example 1

54. The solutions of 𝑥 4 −14 𝑥 2 +45=0 are
{−3, − 5 , 5 , 3}

55. Solving Polynomial Inequalities

56. Use the Rational Root Theorem and the Quadratic Formula to solve a polynomial.
With inequalities, we can use a “test point” to determine the validity of a value.
Things to remember…

57. Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0
Steps
Example
Set the polynomial equal to zero.
Solve the polynomial.
𝑥 3 +3 𝑥 2 −𝑥−3=0
𝑥={−3, 1, −1}

58. Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0
Use the solutions to break up the number line. -3 -1 1

59. Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0
Plug a “test point” from each section of the number in to the original inequality to test the validity of the value. -5 -3 -2 -1 1 2

60. Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0
Use the information from step 4 to write the solution to the inequality in interval notation.
−3, −1 𝑎𝑛𝑑 (1, ∞) -5 -3 -2 -1 1 2