1. On a blank piece of paper please complete PG 328 complete #1-12 on the Prerequisite Skills

2. Chapter 5: Polynomials & Polynomial Functions
BIG IDEAS:
Performing operations with polynomials
Solving polynomial equations and finding zeros

3. Lesson 1: Use Properties of Exponents

4. How do you simplify algebraic expressions with exponents?
Essential question
How do you simplify algebraic expressions with exponents?

5. VOCABULARY
Scientific notation: The representat5ion of a number in the form c x 10n where 1≤c<10 and n is an integer.

6. Power of a product property
EXAMPLE 1
Evaluate numerical expressions a.
(– )2
= (– 4)2 (25)2
Power of a product property =
Power of a power property =
= 16,384
Simplify and evaluate power. b.
115
118 –1
118
115 =
Negative exponent property
= 118 – 5
Quotient of powers property
= 113
= 1331
Simplify and evaluate power.

7. Product of powers property
EXAMPLE 3
Simplify expressions
a. b–4b6b7
= b–
= b9
Product of powers property b.
r–2 –3 s3
( r –2 )–3
( s3 )–3 =
Power of a quotient property =
r 6
s–9
Power of a power property
= r6s9
Negative exponent property
c m4n –5
2n–5
= 8m4n –5 – (–5)
Quotient of powers property
= 8m4n0= 8m4
Zero exponent property

8. Power of a product property
EXAMPLE 4
Standardized Test Practice
SOLUTION
(x–3y3)2
x5y6 =
(x–3)2(y3)2
x5y6
Power of a product property
x –6y6
x5y6 =
Power of a power property

9. GUIDED PRACTICE
for Examples 3, 4, and 5
Simplify the expression. Tell which properties of
exponents you used.
5. x–6x5 x3
ANSWER
x2 ; Product of powers property
(7y2z5)(y–4z–1)
7z4 y2
; Product of powers property, Negative exponent property
ANSWER

10. GUIDED PRACTICE
for Examples 3, 4, and 5 7.
s 3 2
t–4
s6t8
ANSWER
; Power of a power property, Negative exponent property 8.
x4y–
x3y6 x3
y24
; Quotient of powers property, Power of a Quotient property, Negative exponent property
ANSWER

11. How do you simplify algebraic expressions with exponents.
Essential question
How do you simplify algebraic expressions with exponents.
Use the properties of exponents to rewrite it
with only positive exponents

12. Simplify the expression (-3x3) (5x)

13. Lesson 3: Add, Subtract and Multiply Polynomials

14. What are the special product patterns?
Essential question
What are the special product patterns?

15. VOCABULARY
Like Terms: Terms that have the same variable parts. Constant terms are also like terms.

16. EXAMPLE 1
Add polynomials vertically and horizontally
a. Add 2x3 – 5x2 + 3x – 9 and x3 + 6x in a vertical format.
SOLUTION
a x3 – 5x2 + 3x – 9
+ x3 + 6x
3x3 + x2 + 3x + 2

17. EXAMPLE 1
Add polynomials vertically and horizontally
b. Add 3y3 – 2y2 – 7y and –4y2 + 2y – 5 in a horizontal
format.
(3y3 – 2y2 – 7y) + (–4y2 + 2y – 5)
= 3y3 – 2y2 – 4y2 – 7y + 2y – 5
= 3y3 – 6y2 – 5y – 5

18. EXAMPLE 2
Subtract polynomials vertically and horizontally
a. Subtract 3x3 + 2x2 – x + 7 from 8x3 – x2 – 5x + 1 in a vertical format.
SOLUTION
a. Align like terms, then add the opposite of the
subtracted polynomial.
8x3 – x2 – 5x + 1
– (3x3 + 2x2 – x + 7)
8x3 – x2 – 5x + 1
+ – 3x3 – 2x2 + x – 7
5x3 – 3x2 – 4x – 6

19. EXAMPLE 2
Subtract polynomials vertically and horizontally
b. Subtract 5z2 – z + 3 from 4z2 + 9z – 12 in a horizontal
format.
Write the opposite of the subtracted polynomial,
then add like terms.
(4z2 + 9z – 12) – (5z2 – z + 3) = 4z2 + 9z – 12 – 5z2 + z – 3
= 4z2 – 5z2 + 9z + z – 12 – 3
= –z2 + 10z – 15

20. GUIDED PRACTICE
for Examples 1 and 2
Find the sum or difference.
1. (t2 – 6t + 2) + (5t2 – t – 8)
ANSWER
6t2 – 7t – 6
2. (8d – 3 + 9d3) – (d3 – 13d2 – 4)
ANSWER
8d3 + 13d2 + 8d + 1

21. Multiply –2y2 + 3y – 6 by –2 . Multiply –2y2 + 3y – 6 by y
EXAMPLE 3
Multiply polynomials vertically and horizontally
a. Multiply –2y2 + 3y – 6 and y – 2 in a vertical format.
b. Multiply x + 3 and 3x2 – 2x + 4 in a horizontal format.
SOLUTION
a –2y2+ 3y – 6
y – 2
4y2 – 6y + 12
Multiply –2y2 + 3y – 6 by –2 .
–2y3 + 3y2 – 6y
Multiply –2y2 + 3y – 6 by y
–2y3 + 7y2 –12y + 12
Combine like terms.

22. EXAMPLE 3
Multiply polynomials vertically and horizontally
b. (x + 3)(3x2 – 2x + 4) = (x + 3)3x2 – (x + 3)2x + (x + 3)4
= 3x3 + 9x2 – 2x2 – 6x + 4x + 12
= 3x3 + 7x2 – 2x + 12

23. EXAMPLE 4
Multiply three binomials
Multiply x – 5, x + 1, and x + 3 in a horizontal format.
(x – 5)(x + 1)(x + 3) = (x2 – 4x – 5)(x + 3)
= (x2 – 4x – 5)x + (x2 – 4x – 5)3
= x3 – 4x2 – 5x + 3x2 – 12x – 15
= x3 – x2 – 17x – 15

24. Sum and difference Square of a binomial Cube of a binomial EXAMPLE 5
Use special product patterns
a. (3t + 4)(3t – 4)
= (3t)2 – 42
Sum and difference
= 9t2 – 16
b. (8x – 3)2
= (8x)2 – 2(8x)(3) + 32
Square of a binomial
= 64x2 – 48x + 9
c. (pq + 5)3
Cube of a binomial
= (pq)3 + 3(pq)2(5) + 3(pq)(5)2 + 53
= p3q3 + 15p2q2 + 75pq + 125

25. GUIDED PRACTICE
for Examples 3, 4 and 5
Find the product.
3. (x + 2)(3x2 – x – 5)
ANSWER
3x3 + 5x2 – 7x – 10
4. (a – 5)(a + 2)(a + 6)
(a3 + 3a2 – 28a – 60)
ANSWER
5. (xy – 4)3
ANSWER
x3y3 – 12x2y2 + 48xy – 64

26. What are the special product patterns?
Essential question
What are the special product patterns?
(a+b)(a-b) = a2 – b2
(a+b)2= a2 + 2ab + b2
(a-b)2= a2 - 2ab + b2
(a+b)3= a3 + 3a2b + 3ab2 + b3
(a-b)3= a3 - 3a2b + 3ab2 - b3

27. Multiply the polynomials: (x+2) (x+3) (2x – 1) (2x + 1)

28. Lesson 4: Factor and Solve Polynomial Equations

29. How can you solve a higher-degree polynomial equation?
Essential question
How can you solve a higher-degree polynomial equation?

30. VOCABULARY
Factored Completely: A factorable polynomial with integer coefficients is factored completely if it is written as a product of un-factorable polynomials with integer coefficients
Factor by grouping: To factor a polynomial with four terms by grouping, factor common monomials from pairs of terms, then look for a common binomial factor
Quadratic Form: The form au2 + bu + c, where u is any expression in x.

31. Perfect square trinomial
EXAMPLE 1
Find a common monomial factor
Factor the polynomial completely.
a. x3 + 2x2 – 15x
= x(x2 + 2x – 15)
Factor common
monomial.
= x(x + 5)(x – 3)
Factor trinomial.
b. 2y5 – 18y3
= 2y3(y2 – 9)
Factor common
monomial.
= 2y3(y + 3)(y – 3)
Difference of two
squares
c. 4z4 – 16z3 + 16z2
= 4z2(z2 – 4z + 4)
Factor common
monomial.
= 4z2(z – 2)2
Perfect square trinomial

32. Factor common monomial.
EXAMPLE 2
Factor the sum or difference of two cubes
Factor the polynomial completely.
a. x3 + 64
= x3 + 43
Sum of two cubes
= (x + 4)(x2 – 4x + 16)
b z5 – 250z2
= 2z2(8z3 – 125)
Factor common monomial.
= 2z2 (2z)3 – 53
Difference of two cubes
= 2z2(2z – 5)(4z2 + 10z + 25)

33. GUIDED PRACTICE
for Examples 1 and 2
Factor the polynomial completely.
x3 – 7x2 + 10x
ANSWER
x( x – 5 )( x – 2 )
y5 – 75y3
ANSWER
3y3(y – 5)(y + 5 )
b b2
ANSWER
2b2(2b + 7)(4b2 –14b + 49)
4. w3 – 27
ANSWER
(w – 3)(w2 + 3w + 9)

34. Distributive property
EXAMPLE 3
Factor by grouping
Factor the polynomial x3 – 3x2 – 16x + 48 completely.
x3 – 3x2 – 16x + 48
= x2(x – 3) – 16(x – 3)
Factor by grouping.
= (x2 – 16)(x – 3)
Distributive property
= (x + 4)(x – 4)(x – 3)
Difference of two
squares

35. Write as difference of two squares. Difference of two squares
EXAMPLE 4
Factor polynomials in quadratic form
Factor completely: (a) 16x4 – 81 and (b) 2p8 + 10p5 + 12p2.
a x4 – 81
= (4x2)2 – 92
Write as difference
of two squares.
= (4x2 + 9)(4x2 – 9)
Difference of two
squares
= (4x2 + 9)(2x + 3)(2x – 3)
Difference of two
squares
Factor common
monomial.
b. 2p8 + 10p5 + 12p2
= 2p2(p6 + 5p3 + 6)
Factor trinomial in
quadratic form.
= 2p2(p3 + 3)(p3 + 2)

36. GUIDED PRACTICE
for Examples 3 and 4
Factor the polynomial completely.
x3 + 7x2 – 9x – 63
ANSWER
(x + 3)(x – 3)(x + 7)
g4 – 625
(4g2 + 25)(2g + 5)(2g – 5)
ANSWER
t6 – 20t4 + 24t2
4t2(t2 – 3)(t2 – 2 )
ANSWER

37. How can you solve a higher-degree polynomial equation?
Essential question
How can you solve a higher-degree polynomial equation?
Factor the polynomial completely and use the
zero product property

38. A company’s income is modeled by the function P = 22x2 – 571 x
A company’s income is modeled by the function P = 22x2 – 571 x. What is the value of P when x = 200?

39. Lesson 5: Apply the Remainder and Factor Theorem

40. Essential question
If you know one zero of a polynomial function, how can you determine another zero?

41. VOCABULARY
Polynomial long division: A method used to divide polynomials similar to the way you divide numbers
Synthetic division: A method used to divide a polynomial by a divisor of the form x - k

42. EXAMPLE 1
Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 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. This gives the next term of the quotient.

43. ) quotient Multiply divisor by 3x4/x2 = 3x2
EXAMPLE 1
Use polynomial long division
3x2 + 4x – 3
quotient
x2 – 3x + 5
3x4 – 5x3 + 0x2 + 4x – 6 )
Multiply divisor by 3x4/x2 = 3x2
3x4 – 9x3 + 15x2
Subtract. Bring down next term.
4x3 – 15x x
4x3 – 12x2 + 20x
Multiply divisor by 4x3/x2 = 4x
Subtract. Bring down next term.
–3x2 – 16x – 6
–3x2 + 9x – 15
Multiply divisor by – 3x2/x2 = – 3
–25x + 9
remainder

44. EXAMPLE 1
Use polynomial long division
3x4 – 5x3 + 4x – 6
x2 – 3x + 5
= 3x2 + 4x – 3 +
–25x + 9
ANSWER
CHECK
You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend.
(3x2 + 4x – 3)(x2 – 3x + 5) + (–25x + 9)
= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6

45. ) quotient Multiply divisor by x3/x = x2. Subtract.
EXAMPLE 2
Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x
quotient
x – 2
x3 + 5x2 – 7x )
x3 – 2x2
Multiply divisor by x3/x = x2.
7x2 – 7x
Subtract.
7x2 – 14x
Multiply divisor by 7x2/x = 7x.
7x + 2
Subtract.
7x – 14
Multiply divisor by 7x/x = 7. 16
remainder
ANSWER
x3 + 5x2 – 7x +2
x – 2
= x2 + 7x + 7 + 16

46. GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
(2x4 + x3 + x – 1) (x2 + 2x – 1)
(2x2 – 3x + 8) +
–18x + 7
x2 + 2x – 1
ANSWER
(x3 – x2 + 4x – 10)  (x + 2)
(x2 – 3x + 10) +
–30
x + 2
ANSWER

47. EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
– –
– –21
2 – –16
2x3 + x2 – 8x + 5
x + 3
= 2x2 – 5x + 7 – 16
ANSWER

48. EXAMPLE 4
Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that
x + 2 is a factor.
SOLUTION
Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other factors.
– –4 – –16 –
3 – –

49. Write original polynomial.
EXAMPLE 4
Factor a polynomial
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16
Write original polynomial.
= (x + 2)(3x2 – 10x – 8)
Write as a product of two
factors.
= (x + 2)(3x + 2)(x – 4)
Factor trinomial.

50. GUIDED PRACTICE
for Examples 3 and 4
Divide using synthetic division.
3. (x3 + 4x2 – x – 1)  (x + 3)
x2 + x – 4 + 11
x + 3
ANSWER
4. (4x3 + x2 – 3x + 7)  (x – 1)
4x2 + 5x + 2 + 9
x – 1
ANSWER

51. GUIDED PRACTICE
for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a factor.
f (x) = x3 – 6x2 + 5x + 12
ANSWER
(x – 4)(x –3)(x + 1)
f (x) = x3 – x2 – 22x + 40
ANSWER
(x – 4)(x –2)(x +5)

52. Use synthetic division and factor the results
Essential question
If you know one zero of a polynomial function, how can you determine another zero?
Use synthetic division and factor the results
to find the other zeros.