Transcript Intro to Polynomial Factorizations and Equations

Copyright © 2012 Pearson Education, Inc.
Slide 6- 1
6.1
Introduction to
Polynomial
Factorizations
and Equations
■
■
■
Graphical Solutions
The Principle of Zero Products
Terms with Common Factors
■ Factoring by Grouping
■ Factoring and Equations
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Graphical Solutions
Whenever two polynomials are set equal to each other,
the result is a polynomial equation.
In this section we learn how to solve polynomial
equations both graphically and algebraically.
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Slide 6- 3
Example
Solve: x2 = 4x.
Solution
Intersect Method: We can find the real-number
solutions of a polynomial equation by finding the
points of intersection of two graphs.
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Slide 6- 4
continued
Zero Method: Rewrite the equation so that one side is
0 and then find the x-intercepts of one graph or the
zeros of the function.
Rewrite the equation so that one side is 0:
x2 = 4x
x2 – 4x = 0
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Slide 6- 5
Zeros and Roots
The x-values for which a function f(x) is 0 are called
the zeros of the function.
The x-values for which an equation such as f(x) = 0 is
true are called the roots of the equation.
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Slide 6- 6
Example
Find the zeros of the function given by
f(x) = x3 – 2x2 – 5x + 6.
Solution
Graph the equation, choosing a window that shows the
x-intercepts of the graph. This may require several
attempts.
To find the zeros
use the ZERO
option from the
CALC menu.
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Slide 6- 7
continued
f(x) = x3 – 2x2 – 5x + 6
To find the zeros use the ZERO
option from the CALC menu.
Use the same
procedure for
the other two
zeros.
The zeros are
2, 1 and 3.
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Slide 6- 8
The Principle of Zero Products
The Principle of Zero Products
For any real numbers a and b: If ab = 0, then a = 0
or b = 0.
If a = 0 or b = 0, then ab = 0.
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Slide 6- 9
Example
Solve: (x – 4)(x + 3) = 0.
Solution
According to the principle of zero
(4, 0)
products, at least one factor
(3, 0)
must be 0.
x–4=0
or x + 3 = 0
x=4
or
x = 3
For 4:
For 3:
(x – 4)(x + 3) = 0
(x – 4)(x + 3) = 0
(4 – 4)(4 + 3) = 0
(3 – 4)(3 + 3) = 0
0(7) = 0
0(7) = 0
0 = 0 TRUE
0 = 0 TRUE
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Slide 6- 10
Factoring
To factor a polynomial is to find an equivalent
expression that is a product. An equivalent expression
of this type is called a factorization of the
polynomial.
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Slide 6- 11
Terms with Common Factors
When factoring a polynomial, we look for factors
common to every term and then use the distributive
law.
Multiply
Factor
4x(x2 + 3x  4)
= 4xx2 + 4x3x  4x4
= 4x3 + 12x2  16x
4x3 + 12x2  16x
= 4xx2 + 4x3x  4x4
= 4x(x2 + 3x  4)
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Slide 6- 12
Example
Factor: 28x6 + 32x3.
Solution
The prime factorization of 28x6 is
227xxxxxx
The prime factorization of 32x3 is
22222xxx
The largest common factor is 2  2  x  x  x or 4x3.
28x6 + 32x3 = 4x3  7x3 + 4x3  8
= 4x3(7x3 + 8)
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Slide 6- 13
Example Factor: 12x5  21x4 + 24x3
Solution
The prime factorization of 12x5 is
223xxxxx
The prime factorization of 21x4 is
37xxxx
The prime factorization of 24x3 is
2223xxx
The largest common factor is 3  x  x  x or 3x3.
12x5  21x4 + 24x3 = 3x3  4x2  3x3  7x + 3x3  8
= 3x3(4x2  7x + 8)
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Slide 6- 14
Tips for Factoring
1. Factor out the largest common factor, if one exists.
2. The common factor multiplies a polynomial with the
same number of terms as the original polynomial.
3. Factoring can always be checked by multiplying.
Multiplication should yield the original polynomial.
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Slide 5- 15
Factoring by Grouping
Sometimes algebraic expressions contain a common
factor with two or more terms.
Example Factor x2(x + 2) + 3(x + 2).
Solution The binomial (x + 2) is a factor of both
x2(x + 2) and 3(x + 2). Thus, x + 2 is a common factor.
x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3
= (x + 2)(x2 + 3)
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Slide 6- 16
Example
Write an equivalent expression by factoring.
a) 3x3 + 9x2 + x + 3
b) 9x4 + 6x  27x3  18
Solution
a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3)
= 3x2(x + 3) + 1(x + 3)
= (x + 3)(3x2 + 1)
Don’t forget to
include the 1.
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Slide 6- 17
Example continued
b) 9x4 + 6x  27x3  18
= (9x4 + 6x) + (27x3  18)
= 3x(3x3 + 2) + (9)(3x3 + 2)
= (3x3 + 2)(3x  9)
= (3x3 + 2)3(x  3)
= 3(3x3 + 2)(x  3)
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Slide 6- 18
Factoring out ‒1
b – a = ‒1(a – b) = ‒(a – b)
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Slide 5- 19
Example Factor: ax – bx + by – ay.
Solution
ax – bx + by – ay = (ax – bx) + (by – ay)
= x(a – b) + y(b – a)
= x(a – b) + y(‒1)(a – b)
= x(a – b) – y(a – b)
= (a – b)(x – y)
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Slide 6- 20
Factoring and Equations
Example Solve: 7x2 = 35x.
Solution
Use the principle of zero products if there is a 0 on one
side of the equation and the other side is in factored
form.
7x2 = 35x
7x2 – 35x = 0
Subtracting 35x. One side is now 0.
7x(x – 5) = 0
Factoring
7x = 0
or x – 5 = 0 Use the principle of zero products
x=0
or
x=5
We check by substitution or graphically.
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Slide 6- 21
To Use the Principle of Zero Products
1. Write an equivalent equation with 0 on one side,
using the addition principle.
2. Factor the nonzero side of the equation.
3. Set each factor that is not a constant equal to 0.
4. Solve the resulting equations.
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