Download Chapter 5-5

Chapter 5
Polynomials
and Polynomial
Functions
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-1
1
Chapter Sections
5.1 – Addition and Subtraction of Polynomials
5.2 – Multiplication of Polynomials
5.3 – Division of Polynomials and Synthetic
Division
5.4 – Factoring a Monomial from a Polynomial
and Factoring by Grouping
5.5 – Factoring Trinomials
5.6 – Special Factoring Formulas
5.7-A General Review of Factoring
5.8- Polynomial Equations
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-2
2
§ 5.5
Factoring
Trinomials
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-3
3
Factoring Trinomials
Recall that factoring is the reverse process of
multiplication. Using the FOIL method, we
can show that
(x – 3)(x – 8) = x2 – 11x + 24.
Therefore x2 – 11x + 24 = (x – 3)(x – 8).
Note that this trinomial results in the
product of two binomials whose first
term is x and second term is a number
(including its sign).
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-4
4
Factoring Trinomials
Factoring any polynomial of the form x2 + bx + c
will result in a pair of binomials:
x2 + bx + c = (x +?)(x +?)
Numbers go here.
L
F
F
O
I
L
(7x + 3)(2x + 4) = 14x2 + 28x + 6x + 12
I
= 14x2 + 34x
+ 12
O
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-5
5
Factoring Trinomials
1. Find two numbers (or factors) whose
product is c and whose sum is b.
2. The factors of the trinomial will be of the
form
(x + one number) (x + second number)
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-6
6
Examples
a.) Factor x2 - x – 12.
a = 1, b = -1, c = -12. We must find two
numbers whose product is c, which is -12,
and whose sum is b, which is -1. We begin by
listing the factors of -12, trying to find a pair
whose sum is -1.
continued
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-7
7
Examples
Factors of -12
(1)(-12)
(2)(-6)
(3)(-4)
(4)(-3)
(6)(-2)
(12)(1)
Sum of Factors
1 + (-12) = -11
2 + (-6) = -4
3 + (-4) = -1
4 + (-3) = 1
6 + (-2) = 4
12 + (-1) = 11
The numbers we are seeking are 3 and -4 because
their product is -12 and their sum is -1.
x2 - x – 12 = (x + 3)(x – 4)
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-8
8
Factor out a Common Factor
The first step when factoring any trinomial is to
determine whether all three terms have a common
factor. If so, factor out the GCF.
Example Factor 3x4 – 6x3 – 72x2. The factor 3x2 is
common to all three terms. Factor it out first.
= 3x2(x2 – 2x – 24)
We find that -6 and 4 are the factors to the
trinomial in the parentheses. Therefore, 3x4 – 6x3 –
72x2 = 3x2(x – 6)(x + 4)
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-9
9
Trial and Error Method
1. Write all pairs of factors of the coefficient
of the squared term, a.
2. Write all pairs of factors of the constant, c.
3. Try various combinations of these factors
until the correct middle term, bx, is found.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-10
10
Trial and Error Method
Example: Factor 3t2 – 13t + 10.
There is no factor common to all three terms.
Next we determine that a is 3 and the only
factors of 3 and 1 are 1 and 3. Therefore we
write, 3t2 – 13t + 10 = (3t )(t )
Next, we look for the factors that give us
the correct middle term, - 13t.
Continued.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-11
11
Trial and Error Method
Example continued:
Possible Factors
Sum of the Products of the
Inner and Outer Terms
(3t – 1)(t – 10)
-31t
(3t – 10)(t – 1)
-13t
(3t – 2)(t – 5)
-17t
(3t – 5)(t – 2)
-11t
Thus, 3t2 – 13t + 10 = (3t – 10)(t – 1).
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-12
12
Factor Trinomials of the Form ax2 + bx + c, a ≠ 1,
Using Grouping
To Factor Trinomials of the Form ax2 + bx + c, a ≠ 1,
Using Grouping
1. Find two numbers whose product is a · c and whose
sum is b.
2. Rewrite the middle term, bx, using the numbers found
in step 1.
3. Factor by grouping.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-13
13
Factor Trinomials Using Substitution
Sometimes a more complicated trinomial can be
factored by substituting one variable for another.
Example Factor 3z4 – 17z2 – 28.
Let x = z2. Then the trinomial can be written
3z 4  17 z 2  28  3( z 2 ) 2  17 z 2  28
 3x 2  17 x  28
 (3x  4)( x  7)
Now substitute z2 for x.
 (3z 2  4)( z 2  7)
Thus, 3z 4  17 z 2  28  (3z 2  4)( z 2  7).
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 5-14
14