MTH 065
Elementary Algebra II
Chapter 6 – Polynomial Factorizations
and Equations
Section 6.1 – Introduction to Polynomial
Factorizations and Equations
Polynomial Equations

Definition


Standard Form




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An equation that states that two
polynomials are equal to each other.
p(x) = 0
Combine like terms
Terms from highest to lowest degree
Leading coefficient positive
Degree of a Polynomial Equation

Degree of the leading coefficient when in
standard form.
Polynomial Equations

Solution

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Solution Set


A value for the variable that makes the
equation true.
Set of all solutions.
Fundamental Theorem of Algebra

An nth degree polynomial equation will
have at MOST n solutions.
Polynomial Equations – Examples

2x – 5 = 13
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x2 + 2x = 5x – 2
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Standard form: 2x – 18 = 0
Degree 1 (linear)
Solution: x = 9
Standard form: x2 – 3x + 2 = 0
Degree 2 (quadratic)
Solutions: x = 1, 2
x3 + 3 = 5x – x2
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
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Standard form: x3 + x2 – 5x + 3 = 0
Degree 3 (cubic)
Solutions: 1, –3
Graphical Solutions
1.
Put the equation in standard form.

2.
Graph the function p(x).

3.
p(x) = 0
Use a graphics calculator or software.
Determine the x-intercepts.

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Where the graph intersects the x-axis.
Values of x where p(x) equals 0.
These are the solutions.
NOTE: The values of x where p(x) = 0 are called …
zeros of p(x)
&
roots of p(x) = 0
Graphical Solutions – Example 1

2x – 5 = 13


2x – 18 = 0
Solution


aka: zeros or roots
x=9











Graphical Solutions – Example 2

x2 + 2x = 5x – 2


x2 – 3x + 2 = 0
Solutions


aka: zeros or roots
x = 1, 2




Graphical Solutions – Example 3

x3 + 3 = 5x – x2


x3 + x2 – 5x + 3 = 0
Solutions



aka: zeros or roots
x = 1, –3








Factored Polynomials
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
A polynomial written as a product of
two or more polynomials.
Examples:

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x2 – 3x + 2 = (x – 1)(x – 2)
x3 + x2 – 5x + 3 = (x – 1)2(x + 3)
The Principle of Zero Products


If ab = 0, then
a = 0,
b = 0,
or both are zero.
Examples:


If 5x = 0,
then x = 0.
If 3(x – 1) = 0,
then x – 1 = 0
(i.e. x = 1)
Solving Factored
Polynomial Equations
NOTES


One side of the equation MUST be zero.
The other side must be factored.
1.
Set each factor equal to zero.
2.
Solve each simpler equation.
3.
Solution set is all of the solutions.
Solving Factored
Polynomial Equations – Example 1

(x – 1)(x – 2) = 0

x–1=0  x=1

x–2=0  x=2

Solutions: x = 1, 2
Solving Factored
Polynomial Equations – Example 2

(x – 1)2(x + 3) = 0

x–1=0  x=1

x + 3 = 0  x = –3

Solutions: x = 1, –3
Solving Factored
Polynomial Equations – Example 3

7x(3x – 2)(5x + 1) = 0

x=0

3x – 2 = 0  x = 2/3

5x + 1 = 0  x = –1/5

Solutions: x = 0, 2/3, –1/5
Solving Polynomial Equations
by Factoring

Using the “Principle of Zero Products”
p(x) = 0 can be solved …
2.
The primary topic
of this chapter!
Set each factor equal to zero
3.
Solve these simpler equations
1.
Factor p(x)
NOTE: If the equation is not originally in
the form p(x) = 0, first determine this
equivalent standard form.
Factoring – Step 1
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Distributive Property



ax + ay = a(x + y)
ax – ay = a(x – y)
Factoring Out Common Factors

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Common
Factors
If ALL of the terms of a polynomial
have a common factor, that common
factor may be “factored out.”
Example:

5x2 – 15x = 5x(x – 3)
Factoring – Step 1
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Note: If the leading coefficient is
negative, it is common to factor out
the negative.
Examples:
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Common
Factors
–3x + 21 = –3(x – 7)
–x2 + 15x = –x(x – 15)
–x3 – 7 = –(x3 + 7)
–6x2 – 2x = –2x(3x + 1)
Notice that the sign of each term is
changed when factoring out a
negative.
Factoring – Step 2
Factoring
by Grouping
Note: This method is used with 4-term polynomials.
x2 + 5x + 2x + 10
1.
Make two groups of two terms.
(x2 + 5x) + (2x + 10)
2.
Factor out common factors from each pair.
x(x + 5) + 2(x + 5)
3.
If the remaining polynomials are the
same, factor out the polynomial.
(x + 5)(x + 2)
NOTE: Be careful with negative terms!
Clarifying Terminology!!!!!

Expressions can
be factored.


2x3 – 12x2
= 2x2(x – 6)
Equations can
be solved.

2x3 – 12x2 = 0
x = 0, 6

Sometimes, an
expression in an
equation is factored
in order to solve the
equation.


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2x3 – 12x2 = 0
2x2(x – 6) = 0
x = 0 or x – 6 = 0
x = 0, 6