Transcript Factoring Trinomials 1

Factoring Polynomials and
Solving Equations
by Factoring
Copyright © Cengage Learning. All rights reserved.
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Section
5.3
Factoring Trinomials with a
Leading Coefficient of 1
Copyright © Cengage Learning. All rights reserved.
Objectives
1 Factor a trinomial of the form
ax2 + bx + c, a = 1, by trial and error.
2 Factor a trinomial containing a negative
greatest common factor.
3 Identify a prime trinomial.
4 Factor a polynomial completely.
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Objectives
5 Factor a trinomial of the form
ax2 + bx + c, a = 1, by grouping (ac method).
6 Factor a perfect-square trinomial.
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Factoring Trinomials with a Leading Coefficient of 1
We now discuss how to factor trinomials of the form
x2 + bx + c, where the coefficient of x2 is 1 and there are no
common factors.
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1. Factor a trinomial of the form
ax2 + bx + c, a = 1 by trial and error
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Factor a trinomial of the form ax2 + bx + c, a = 1 by trial and error
The product of two binomials is often a trinomial. For
example,
(x + 3)(x + 3) = x2 + 6x + 9
and
(x – 3y)(x + 4y) = x2 + xy – 12y2
For this reason, we should not be surprised that many
trinomials factor into the product of two binomials.
To develop a method for factoring trinomials, we multiply
(x + a) and (x + b).
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Factor a trinomial of the form ax2 + bx + c, a = 1 by trial and error
(x + a)(x + b) = x2 + bx + ax + ab
Use the distributive property.
= x2 + ax + bx + ab
Apply the commutative
property of addition.
= x2 + (a + b)x + ab
Factor the GCF,
x, out of ax + bx.
From the result, we can see that
• the first term is the product of x and x.
• the coefficient of the middle term is the sum of a and
b, and
• the last term is the product of a and b.
We can use these facts to factor trinomials with leading
coefficients of 1.
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Example 1
Factor: x2 + 5x + 6
Solution:
To factor this trinomial, we will write it as the product of two
binomials. Since the first term of the trinomial is x2, the first
term of each binomial factor must be x because x  x = x2.
To fill in the following blanks, we must find two integers
whose product is +6 and whose sum is +5.
x2 + 5x + 6 = (x
) (x
)
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Example 1 – Solution
cont’d
The positive factorizations of 6 and the sums of the factors
are shown in the following table.
The last row contains the integers +2 and +3, whose
product is +6 and whose sum is +5. So we can fill in the
blanks with +2 and +3.
x2 + 5x + 6 = (x + 2)(x + 3)
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Example – Solution
cont’d
To check the result, we verify that (x + 2) times (x + 3)
is x2 + 5x + 6.
(x + 2)(x + 3) = x2 + 3x + 2x + 2  3
= x2 + 5x + 6
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Factor a trinomial of the form ax2 + bx + c, a = 1 by trial and error
Comment
In Example 1, the factors can be written in either order due
to the commutative property of multiplication. An equivalent
factorization is x2 + 5x + 6 = (x + 3)(x + 2).
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2.
Factor a trinomial containing a negative
greatest common factor
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Factor a trinomial containing a negative greatest common factor
When the coefficient of the first term is –1, we begin by
factoring out –1.
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Example 6
Factor: –x2 + 2x + 15.
Solution:
We factor out –1 and then factor the trinomial.
–x2 + 2x + 15 = –(x2 – 2x – 15)
= –(x – 5)(x + 3)
Factor out –1.
Factor x2 – 2x – 15.
Check:
–(x – 5)(x + 3) = –(x2 + 3x – 5x – 15)
= –(x2 – 2x – 15)
= –x2 + 2x + 15
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3.
Identify a prime trinomial
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Identify a prime trinomial
If a trinomial cannot be factored using only rational
coefficients, it is called a prime polynomial over the set of
rational numbers.
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Example 7
Factor: x2 + 2x + 3
Solution:
To factor the trinomial, we must find two integers whose
product is +3 and whose sum is 2.
The possible factorizations of 3 and the sums of the factors
are shown in the table.
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Example 7 – Solution
cont’d
Since two integers whose product is +3 and whose sum
is +2 do not exist, x2 + 2x + 3 cannot be factored.
It is a prime trinomial.
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4.
Factor a polynomial completely
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Example
Factor: –3ax2 + 9a – 6ax
Solution:
We write the trinomial in descending powers of x and factor
out the common factor of –3a.
–3ax2 + 9a – 6ax = –3ax2 – 6ax + 9a
= –3a(x2 + 2x – 3)
Then we factor the trinomial x2 + 2x – 3.
–3ax2 + 9a – 6ax = –3a(x + 3)(x – 1)
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Example – Solution
cont’d
Check:
–3a(x + 3)(x – 1) = –3a(x2 + 2x – 3)
= –3ax2 – 6ax + 9a
= –3ax2 + 9a – 6ax
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2 + bx + c,
Factor
a
trinomial
of
the
form
ax
5.
a = 1 by grouping (ac method)
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Factor a trinomial of the form ax2 + bx + c, a = 1 by grouping (ac method)
An alternate way of factoring trinomials of the form
ax2 + bx + c, a = 1 uses the technique of factoring by
grouping, sometimes referred to as the ac method.
For example, to factor x2 + x – 12 by grouping, we proceed
as follows:
1. Determine the values of a and c (a = 1 and c = –12) and
find ac:
(1)(–12) = –12
This number is called the key number.
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Factor a trinomial of the form ax2 + bx + c, a = 1 by grouping (ac method)
2. Find two factors of the key number –12 whose sum is
b = 1.
Two such factors are +4 and –3.
+4(–3) = –12
and
+4 + (–3) = 1
3. Use the factors +4 and –3 as the coefficients of two
terms to be placed between x2 and –12 to replace x.
x2 + x – 12 = x2 + 4x – 3x – 12
x = +4x – 3x
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Factor a trinomial of the form ax2 + bx + c, a = 1 by grouping (ac method)
4. Factor the right side of the previous equation by
grouping.
x2 + 4x – 3x – 12 = x(x + 4) – 3(x + 4)
= (x + 4)(x – 3)
Factor x out of (x2 + 4x)
and –3 out of (–3x – 12).
Factor out (x + 4).
Check this factorization by multiplication.
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Example
Factor y2 + 7y + 10 by grouping.
Solution:
We note that this equation is in the form y2 + by + c, with
a = 1, b = 7, and c = 10.
First, we determine the key number ac:
ac = 1(10) = 10
Then, we find two factors of 10 whose sum is b = 7. Two
such factors are +2 and +5.
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Example 10 – Solution
cont’d
We use these factors as the coefficients of two terms
whose sum is 7y.
y2 + 7y + 10 = y2 + 2y + 5y + 10
7y = +2y + 5y
Finally, we factor the right side of the previous equation by
grouping.
y2 + 2y + 5y + 10 = y(y + 2) + 5(y + 2)
= (y + 2)(y + 5)
Factor out y from (y2 + 2y) and
factor out 5 from (5y + 10).
Factor out (y + 2).
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6. Factor a perfect-square trinomial
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Factor a perfect-square trinomial
We have discussed the following special-product
relationships used to square binomials.
1. (x + y)2 = x2 + 2xy + y2
2. (x – y)2 = x2 – 2xy + y2
These relationships can be used in reverse order to factor
special trinomials called perfect-square trinomials.
Perfect-Square Trinomials
1. x2 + 2xy + y2 = (x + y)2
2. x2 – 2xy + y2 = (x – y)2
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Factor a perfect-square trinomial
In words, Formula 1 states that if a trinomial is the square
of one quantity, plus twice the product of the two quantities,
plus the square of the second quantity, it factors into the
square of the sum of the quantities.
Formula 2 states that if a trinomial is the square of one
quantity, minus twice the product of the two quantities, plus
the square of the second quantity, it factors into the square
of the difference of the quantities.
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Factor a perfect-square trinomial
The trinomials on the left sides of the previous equations
are perfect-square trinomials because they are the results
of squaring a binomial.
Although we can factor perfect-square trinomials by using
the techniques discussed in this section, we usually can
factor them by inspecting their terms.
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Factor a perfect-square trinomial
For example, x2 + 8x + 16 is a perfect-square trinomial,
because
• The first term x2 is the square of x.
• The last term 16 is the square of 4.
• The middle term 8x is twice the product of x and 4.
Thus,
x2 + 8x + 16 = x2 + 2(x)(4) + 42
= (x + 4)2
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Example
Factor: x2 – 10x + 25
Solution:
x2 – 10x + 25 is a perfect-square trinomial, because
• The first term x2 is the square of x.
• The last term 25 is the square of 5.
• The middle term –10x is the negative of twice the
product of x and 5.
Thus,
x2 – 10x + 25 = x2 – 2(x)(5) + 52
= (x – 5)2
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