On a blank piece of paper please
complete PG 328 complete #1-12 on
the Prerequisite Skills
Chapter 5:
Polynomials & Polynomial
Functions
BIG IDEAS:
1. Performing operations with polynomials
2. Solving polynomial equations and finding
zeros
Lesson 1:
Use Properties of Exponents
ESSENTIAL QUESTION
How do you simplify algebraic
expressions with exponents?
VOCABULARY
• Scientific notation: The representat5ion of a
number in the form c x 10n where 1≤c<10 and n
is an integer.
EXAMPLE 1
a.
Evaluate numerical expressions
(–4 25)2
= (– 4)2
(25)2
Power of a product property
= 16 25
2
Power of a power property
= 16 210
b.
115
118
–1
= 16,384
Simplify and evaluate power.
118
=
115
Negative exponent property
= 118 – 5
Quotient of powers property
= 113
= 1331
Simplify and evaluate power.
EXAMPLE 3
a.
b.
Simplify expressions
b–4b6b7
r–2
s3
= b–4 + 6 + 7
–3
( r –2 )–3
=
( s3 )–3
=
c.
16m4n –5
2n–5
r6
s–9
= b9
Product of powers property
Power of a quotient property
Power of a power property
= r6s9
Negative exponent property
= 8m4n –5 – (–5)
Quotient of powers property
= 8m4n0= 8m4
Zero exponent property
EXAMPLE 4
Standardized Test Practice
SOLUTION
(x–3y3)2
x5y6
=
=
(x–3)2(y3)2
x5y6
x –6y6
x5y6
Power of a product property
Power of a power property
GUIDED PRACTICE
for Examples 3, 4, and 5
Simplify the expression. Tell which properties of
exponents you used.
5.
x–6x5 x3
ANSWER
6.
x2 ; Product of powers property
(7y2z5)(y–4z–1)
ANSWER
7z4
y2
; Product of powers property, Negative exponent
property
for Examples 3, 4, and 5
GUIDED PRACTICE
7.
s3
2
t–4
ANSWER
8.
x4y–2
s6t8
; Power of a power property, Negative
exponent property
3
x3y6
x3
ANSWER
y24
; Quotient of powers property, Power of a
Quotient property, Negative exponent property
ESSENTIAL QUESTION
How do you simplify algebraic
expressions with exponents.
Use the properties of exponents to rewrite it
with only positive exponents
Simplify the expression (-3x3) (5x)
Lesson 3:
Add, Subtract and Multiply
Polynomials
ESSENTIAL QUESTION
What are the special product
patterns?
VOCABULARY
• Like Terms: Terms that have the same variable
parts. Constant terms are also like terms.
EXAMPLE 1
Add polynomials vertically and horizontally
a. Add 2x3 – 5x2 + 3x – 9 and x3 + 6x2 + 11 in a vertical format.
SOLUTION
a.
2x3 – 5x2 + 3x – 9
+ x3 + 6x2
+ 11
3x3 + x2 + 3x + 2
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
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
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
for Examples 1 and 2
GUIDED PRACTICE
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
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.
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
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
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
= (pq)3 + 3(pq)2(5) + 3(pq)(5)2 + 53
= p3q3 + 15p2q2 + 75pq + 125
Cube of a
binomial
for Examples 3, 4 and 5
GUIDED PRACTICE
Find the product.
3. (x + 2)(3x2 – x – 5)
ANSWER
3x3 + 5x2 – 7x – 10
4. (a – 5)(a + 2)(a + 6)
ANSWER
(a3 + 3a2 – 28a – 60)
5. (xy – 4)3
ANSWER
x3y3 – 12x2y2 + 48xy – 64
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
Multiply the polynomials:
a) (x+2) (x+3)
b) (2x – 1) (2x + 1)
Lesson 4:
Factor and Solve Polynomial
Equations
ESSENTIAL QUESTION
How can you solve a higher-degree
polynomial equation?
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.
EXAMPLE 1
Find a common monomial factor
Factor the polynomial completely.
a.
b.
c.
x3 + 2x2 – 15x
2y5 – 18y3
= x(x2 + 2x – 15)
Factor common
monomial.
= x(x + 5)(x – 3)
Factor trinomial.
= 2y3(y2 – 9)
Factor common
monomial.
= 2y3(y + 3)(y – 3)
Difference of two
squares
4z4 – 16z3 + 16z2
= 4z2(z2 – 4z + 4)
Factor common
monomial.
= 4z2(z – 2)2
Perfect square trinomial
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.
16z5 – 250z2
= 2z2(8z3 – 125)
Factor common monomial.
= 2z2 (2z)3 – 53
Difference of two cubes
= 2z2(2z – 5)(4z2 + 10z + 25)
GUIDED PRACTICE
for Examples 1 and 2
Factor the polynomial completely.
1.
x3 – 7x2 + 10x
ANSWER
2.
3y5 – 75y3
ANSWER
3.
3y3(y – 5)(y + 5 )
16b5 + 686b2
ANSWER
4.
x( x – 5 )( x – 2 )
2b2(2b + 7)(4b2 –14b + 49)
w3 – 27
ANSWER
(w – 3)(w2 + 3w + 9)
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
EXAMPLE 4
Factor polynomials in quadratic form
Factor completely: (a) 16x4 – 81 and (b) 2p8 + 10p5 + 12p2.
a.
16x4 – 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
b. 2p8 + 10p5 + 12p2
= 2p2(p6 + 5p3 + 6)
Factor common
monomial.
= 2p2(p3 + 3)(p3 + 2)
Factor trinomial in
quadratic form.
GUIDED PRACTICE
for Examples 3 and 4
Factor the polynomial completely.
5.
x3 + 7x2 – 9x – 63
ANSWER
6.
16g4 – 625
ANSWER
7.
(x + 3)(x – 3)(x + 7)
(4g2 + 25)(2g + 5)(2g – 5)
4t6 – 20t4 + 24t2
ANSWER
4t2(t2 – 3)(t2 – 2 )
ESSENTIAL QUESTION
How can you solve a higher-degree
polynomial equation?
Factor the polynomial completely and use the
zero product property
A company’s income is modeled by the
function P = 22x2 – 571 x. What is the value
of P when x = 200?
Lesson 5:
Apply the Remainder and
Factor Theorem
ESSENTIAL QUESTION
If you know one zero of a
polynomial function, how can you
determine another zero?
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
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.
EXAMPLE 1
Use polynomial long division
3x2 + 4x – 3
x2 – 3x + 5
)3x4 – 5x3 +
3x4
–
9x3
+
quotient
0x2 + 4x – 6
Multiply divisor by 3x4/x2 = 3x2
15x2
4x3 – 15x2 + 4x
4x3 – 12x2 + 20x
–3x2 – 16x – 6
–3x2 + 9x – 15
–25x + 9
Subtract.
Bring down next term.
Multiply divisor by 4x3/x2 = 4x
Subtract.
Bring down next term.
Multiply divisor by – 3x2/x2 = – 3
remainder
EXAMPLE 1
Use polynomial long division
ANSWER
3x4 – 5x3 + 4x – 6
x2 – 3x + 5
= 3x2 + 4x – 3 +
–25x + 9
x2 – 3x + 5
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
EXAMPLE 2
Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x +
x–2
) x3 + 5x2 – 7x +
quotient
7
2
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
ANSWER
x3 + 5x2 – 7x +2
x–2
remainder
= x2 + 7x + 7 +
16
x–2
for Examples 1 and 2
GUIDED PRACTICE
Divide using polynomial long division.
1.
(2x4 + x3 + x – 1)
ANSWER
2.
(2x2 – 3x + 8) +
(x3 – x2 + 4x – 10)
ANSWER
(x2 + 2x – 1)

–18x + 7
x2 + 2x – 1
(x + 2)
(x2 – 3x + 10) +
–30
x+2
EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
–3
2
1
2
ANSWER
–8
5
–6
15
–21
–5
7
–16
2x3 + x2 – 8x + 5
x+3
= 2x2 – 5x + 7 –
16
x+3
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
synthetic division to find the other factors.
–2
–4
3
–28
–6
3
–10
–16
20
–8
16
0
f (–2) = 0. Use
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
= (x + 2)(3x2 – 10x – 8)
= (x + 2)(3x + 2)(x – 4)
Write original polynomial.
Write as a product of two
factors.
Factor trinomial.
for Examples 3 and 4
GUIDED PRACTICE
Divide using synthetic division.
3.
(x3 + 4x2 – x – 1)  (x + 3)
ANSWER
x2 + x – 4 +
11
x+3
4.
(4x3 + x2 – 3x + 7)  (x – 1)
ANSWER
4x2
+ 5x + 2 +
9
x–1
for Examples 3 and 4
GUIDED PRACTICE
Factor the polynomial completely given that x – 4 is a factor.
5.
f (x) = x3 – 6x2 + 5x + 12
ANSWER
6.
(x – 4)(x –3)(x + 1)
f (x) = x3 – x2 – 22x + 40
ANSWER
(x – 4)(x –2)(x +5)
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.