Transcript a x

Exponents,
Polynomials and
Functions
Copyright © Cengage Learning. All rights reserved.
3
3.4
Factoring Polynomials
Copyright © Cengage Learning. All rights reserved.
Objectives

Factor out the GCF.

Factor by grouping.

Factor quadratics using the AC method.

Factor using trial and error.
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Factoring Out the GCF
4
Factoring Out the GCF
We will multiply the two polynomials using the distributive
property.
a(x + y) = ax + ay
When working with polynomials, we sometimes want to
“undo” the distributive property by changing a polynomial
into a product of two or more factors.
The process of changing a polynomial from terms that are
added together to factors that are multiplied together is
called factoring.
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Factoring Out the GCF
Factoring is one method that is used to solve some
polynomial equations, and it has several other uses in
algebra.
The basis of factoring is to take a polynomial and rewrite it
into simpler parts that are multiplied together.
Standard form
x2 + 8x + 15
Factored form
(x + 3)(x + 5)
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Factoring Out the GCF
When the factored form is multiplied out, the expression will
return to the standard form.
(x + 3)(x + 5) = x2 + 5x + 3x + 15
= x2 + 8x + 15
The most basic step in any factoring process is to factor out
the common elements from the terms of an expression.
When we factor out a common element, it does not go
away, but it becomes a factor that is in front of the
remaining expression.
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Factoring Out the GCF
The common elements of an expression are called a
common factor. The terms of the expression 12x + 10 both
are divisible by 2.
Therefore, a 2 can be factored out of the expression and be
put in front as a factor.
12x + 10 = 2(6x + 5)
If we multiply the 2 back into the expression 6x + 5, we
would be right back to the original expression,12x + 10.
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Factoring Out the GCF
The terms in some expressions will have constants,
variables, or both variables and constants in common.
The greatest common factor (GCF) is the largest
common factor shared in common with all the terms of the
polynomial.
When factoring, we will most often factor out the GCF.
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Example 1 – Factoring out the greatest common factor
Factor out the greatest common factor.
a. 5x2 + 20x
b. 15w3z – 9w2z2
c. 2x2 + 6x – 4
d. 5x(x – 8) + 2(x – 8)
e. 3z(z – 5) – (z – 5)
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Example 1 – Solution
a. Both terms in the expression 5x2 + 20x have at least one
x and also are divisible by 5, so we can factor out 5x
from both terms.
5x(x + 4)
The first term of this expression had x2, but one is
factored out, leaving a single x.
b. In the expression 15w3z – 9w2z2, both of the terms have
a w2, have a z, and are divisible by 3. Factoring 3w2z
out, we have
3w2z(5w – 3z)
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Example 1 – Solution
cont’d
c. All of the terms in the polynomial 2x2 + 6x – 4 are
divisible by 2. Because the last term does not have an x
in it, x is not a common factor,
2(x2 + 3x – 2)
d. In the expression 5x(x – 8) + 2(x – 8), both terms have
the expression (x – 8).
(x – 8)(5x + 2)
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Example 1 – Solution
cont’d
Both terms in the expression 3z(z – 5) – (z – 5) have
the expression (z – 5). Remember that the expression
(z – 5) has a “hidden” coefficient of –1. When we factor out
the (z – 5), a – 1 will remain in its place.
(z – 5)(3z – 1)
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Factoring by Grouping
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Factoring by Grouping
Factoring out what is in common will always be the first
step in any factoring process.
In many cases, there will not be anything in common, and
we will move on to other steps. If a polynomial has four
terms, we try a factoring technique called factoring by
grouping.
We will group the first two terms together and factor out the
GCF of those two terms. Also we group the last two terms
and factor out the GCF.
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Factoring by Grouping
We finish by factoring out the GCF from the remaining
expressions.
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Example 2 – Factoring by grouping
Factor by grouping.
a. 2x2 – 8x + 7x – 28
b. 6a2 + 8ab + 15a + 20b
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Example 2(a) – Solution
There are four terms in the expression 2x2 – 8x + 7x – 28,
so we will factor by grouping. There are no common factors
for all four terms. Group the first two terms and the last two
terms, then factor out the GCFs.
2x2 – 8x + 7x – 28 = (2x2 – 8x) + (7x – 28)
Group the first two
and last two terms
= 2x(x – 4) + 7(x – 4)
Factor out the GCFs
2x and 7.
= (x – 4)(2x + 7)
Factor out the GCF
(x – 4)
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Example 2(a) – Solution
cont’d
To check the factorization, we can multiply out the factored
form using the distributive property.
(x – 4)(2x + 7) = 2x2 – 8x + 7x – 28
We got the original expression back, so the factored form
we found is correct.
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Example 2(b) – Solution
cont’d
There are four terms in the expression
6a2 + 8ab + 15a + 20b, so we will factor by grouping.
There are no common factors for all four terms. Group the
first two terms and the last two terms, then factor out the
GCFs.
6a2 + 8ab + 15a + 20b
= (6a2 + 8ab + 15a + 20b)
Group the first two
and last two terms.
= 2a(3a + 4b + 5(3a + 4b)
Factor out the GCFs
2a and 5.
= (3a + 4b)(2a + 5)
Factor out the
GCF (3a + 4b).
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Example 2(b) – Solution
cont’d
To check the factorization, we can multiply out the
factored form using the distributive property.
(3a + 4b)(2a + 5) = 6a2 + 8ab + 15a + 20b
We got the original expression back, so the factored form
we found is correct.
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Factoring Using the AC Method
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Factoring Using the AC Method
One of the most common polynomial functions that we will
work with is called a quadratic. Quadratic functions are
second-degree polynomials of the form
f(x) = ax2 + bx + c
where a, b, and c are real numbers and a ≠ 0.
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Factoring Using the AC Method
Many methods are used to factor quadratics, but we will
concentrate on one called the AC method.
In working with a quadratic in standard form, this method
will provide basic steps that will guide us through the
factoring process.
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Example 3 – Factor using the AC method
Factor the following.
a. x2 + 8x + 15
b. 6x2 + x – 35
c. 4x2 + 7xy + 3y2
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Example 3(a) – Solution
The quadratic x2 + 8x + 15 is in standard form ax2 + bx + c,
so we will use the AC method.
Step 1: Factor out the greatest common factor. The terms
in this quadratic have no common factors.
Step 2: Multiply a and c together. (Do this step off to
the side, in the margin.)
a=1
and
c = 15
so
ac = 1(15) = 15
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Example 3(a) – Solution
cont’d
Step 3: Find factors of ac that sum to b. (Do this off to the
side also.)
List both the positive and negative factors of 15.
ac = 1(15) = 15
b=8
Not equal to b
1
15
1 + 15 = 16
3
5
3+5=8
–1
–15
–1 + (–15) = –16
Not equal to b
–3
–5
–3 + (–5) = –8
Not equal to b
Equal to b
In this list, the factors 3 and 5 will add up to 8.
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Example 3(a) – Solution
cont’d
Step 4: Rewrite the middle (bx) term, using the factors from
step 3.
The factors 3 and 5, found in the previous step, will
be used to rewrite the bx term in the expression.
x2 + 3x + 5x + 15
Step 5: Group and factor out what is in common. If we
group the first two terms together and the last two
terms together.
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Example 3(a) – Solution
cont’d
We can factor out some common factors.
x2 + 3x + 5x + 15
= (x2 + 3x) + (5x + 15)
Group first and last
two terms together.
= x(x + 3) + 5(x + 3)
Factor out x from the first
group and 5 from the
second group.
= (x + 3)(x + 5)
Factor out the (x + 3).
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Example 3(a) – Solution
cont’d
To check the factorization, we can multiply out the
factored form using the distributive property.
(x + 3)(x + 5) = x2 + 5x + 3x + 15
= x2 + 8x + 15
We got the original expression back, so the factored
form we found is correct.
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Example 3(b)– Solution
cont’d
The quadratic 6x2 + x – 35 is in standard form ax2 + bx + c,
so we will use the AC method.
Step 1: Factor out the greatest common factor. The
terms in this quadratic have nothing in common.
Step 2: Multiply a and c together. (Do this step off to the
side, in the margin.)
a = 6 and c = –35
so
ac = 6(–35) = – 210.
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Example 3(b) – Solution
cont’d
Step 3: Find factors of ac that sum to b. (Do this off to the
side also.) List the factors of –210.
Once we have listed the factors of –120, then we
will switch the signs of the factors to come up with
the entire list of factors for –120.
The product must be
negative, so one factor must
be positive and the other
must be negative. Be sure to
switch the signs of the
factors to find all the
factorizations of –120.
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Example 3(b) – Solution
cont’d
In this list, the factors –14 and 15 will add up to 1.
Step 4: Rewrite the middle (bx) term, using the factors
from step 3. The factors –14 and 15 will replace
the b value in the expression.
6x2 –14x + 15x – 35
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Example 3(b) – Solution
cont’d
Step 5: Group and factor out what is in common.
If we group the first two terms together and the
last two terms together, we can factor out some
common factors.
6x2 – 14x + 15x – 35
= (6x2 – 14x) + (15x – 35)
= 2x(3x – 7) + 5(3x – 7)
Group the first and last two
terms together. Factor out2x
from the first group and 5
from the second group.
Factor out the (3x – 7).
= (3x – 7)(2x + 5)
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Example 3(b) – Solution
cont’d
We will check this factorization by multiplying the
factored form out using the distributive property.
(3x – 7)(2x + 5) = 6x2 + 15x – 14x – 35
= 6x2 + x – 35
This is the same expression that we started with, so
the factored form is correct.
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Example 3(c) – Solution
cont’d
The polynomial 4x2 + 7xy + 3y2 has two variables, but
each variable has a squared term and a first-degree
term.
We will factor it using the AC method focusing on the x
variables, and the y variables will follow along the process.
Step 1: Factor out the greatest common factor. The
terms in this polynomial have nothing in common.
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Example 3(c) – Solution
cont’d
Step 2: Multiply a and c together. (Do this step off to
the side, in the margin.)
a = 4 and c = 3 so ac = 4(3) = 12
Step 3: Find factors of ac that sum to b. (Do this off to
the side also.) List the factors of 12.
12
12
1
12
–1 –12
2
6
–2
–6
3
4
–3
–4
In this list, the factors 3 and 4 will add up to 7.
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Example 3(c) – Solution
cont’d
Step 4: Rewrite the middle (bx) term, using the factors
from step 3. The factors 3 and 4 will replace the
b value in the expression.
4x2 + 4xy + 3xy + 3y2
Step 5: Group and factor out what is in common. If we
group the first two terms together and the last
two terms together, we can factor out some
common factors.
4x2 + 4xy + 3xy + 3y2 = (4x2 + 4xy) + (3xy + 3y2)
Group the first two terms and the last two terms.
Factor out 4x from the first group and 3y from the
second group. Factor out the (x + y)
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Example 3(c) – Solution
cont’d
= 4x(x + y) + 3y(x + y )
= (x + y)(4x + 3y)
We will check this factorization by multiplying the
factored form out using the distributive property.
(x + y)(4x + 3y) = 4x2 + 3xy + 4xy + 3y2
= 4x2 + 7xy + 3y2
This is the same expression that we started with, so the
factored form is correct.
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Factoring Using Trial and Error
40
Factoring Using Trial and Error
Another method used to factor quadratics and other
polynomials is trial and error. This method uses our
knowledge of multiplying and adding numbers to help us
guess the factors of the quadratic.
The first step of this method will also be to take out any
common factors. After factoring out anything common, we
will use factors of a and c to try to find the factored form of
the expression.
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Factoring Using Trial and Error
If a = 1 in the standard form, we will have to consider only
the factors of c.
a=1
x2 + 5x + 6
(x + factor of c)(x + factor of c)
(x + 2)(x + 3)
If a does not equal 1, we will have to consider both the
factors of a and c.
a≠1
10x2 + 33x + 20
(factor of a  x + 1 factor of c)(factor of a  x + factor of c)
(2x + 5)(5x + 4)
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Factoring Using Trial and Error
The signs of the factors are another aspect of the factored
form that we will have to be cautious about. If both of the
terms in the original polynomial expression have plus signs,
we will have two plus signs in the factored form.
If the polynomial has one or more negative signs, we will
have at least one negative sign in the factored form.
x2 + 9x + 20
x2 + 2x – 8
x2 – 3x + 2
(x + 5)(x + 4)
(x + 4)(x – 2) (x – 2)(x – 1)
x2 – 4x – 5
(x – 5)(x + 1)
It is best not to attempt to memorize every situation. Learn
to try something, check it, and try again if it does not work.
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Example 6 – Factor using trial and error
Factor the following using trial and error.
a. x2 + 7x + 6
b. x2 – 4x – 12
c. 2x2 + 9x + 4
d. 24x2 – 2x – 15
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Example 6(a) – Solution
cont’d
Because a is 1 in the expression x2 + 7x + 6, we have
to consider only the factors of c. The factors of 6 are
1 and 6 or 2 and 3. Because 1 and 6 add up to 7, the
factorization is
x2 + 7x + 6 = (x + 6)(x + 1)
We check this by multiplying the factored form out.
(x – 6)(x + 1) = x2 + x + 6x + 6
= x2 + 7x + 6
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Example 6(b) – Solution
cont’d
Again, a is 1 in the expression x2 – 4x – 12, so we
consider only the factors of c. The factors of 12 are 1
and 12, 2 and 6, or 3 and 4.
Because we have negatives in the original
expression, we will have one or more negatives in
the factored form.
The factors 2 and 6 can make –4 if the 2 is positive
and the 6 is negative.
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Example 6(b) – Solution
cont’d
This gives us the factored form.
x2 – 4x – 12 = (x – 6)(x + 2)
We check this factored form by multiplying it out.
(x – 6)(x + 2) = x2 + 2x – 6x – 12
= x2 – 4x – 12
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Example 6(c) – Solution
cont’d
Because a is not 1 in the expression 2x2 + 9x + 4, we
have to consider both the factors of a and c. The
factors of 2 are 1 and 2, and the factors of 4 are 1 and
4 or 2 and 2. Using these, we find the factored form.
2x2 + 9x + 4 = (2x + 1)(x + 4)
We check this factored form by multiplying it out.
(2x + 1)(x + 4) = 2x2 + 8x + x + 4
= 2x2 + 9x + 4
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Example 6(d) – Solution
cont’d
Again, a is not 1 in the expression 24x2 – 2x – 15, so
we have to consider the factors of a and c.
The factors of 24 are 1 and 24, 2 and 12, 3 and 8, and
4 and 6. The factors of 15 are 1 and 15 or 3 and 5.
The negatives in the original expression mean that we
will have one or more negatives in the factored form.
24x2 – 2x – 15
Try 1
(2x + 5)(12x – 3)
Try 2
(3x + 5)(8x – 3)
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Example 6(d) – Solution
24x2 – 6x + 60x – 15
24x2 – 9x + 40x – 15
24x2 + 54x – 15
24x2 + 31x – 15
Wrong middle term.
Try 3
cont’d
Wrong middle term.
Try 4
(4x + 1)(6x – 15)
(4x + 3)(6x – 5)
24x2 – 60x + 6x – 15
24x2 – 20x + 18x – 15
24x2 – 54x – 15
24x2 – 2x – 15
Wrong middle term.
This one works!
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Prime Polynomials
51
Prime Polynomials
Some polynomials will not be able to be factored by using
rational numbers. When a polynomial is not factorable, it is
called prime.
When using the AC method, we know that a polynomial is
not factorable if in step 3 there is no combination of factors
for ac that add up to b.
If this happens, we will stop and say that the polynomial is
prime.
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Prime Polynomials
When factoring by trial and error, we must try every
combination of factors of a and c to see whether any of the
combinations work.
If no combination of factors and positive and negative signs
work, we will say the polynomial is not factorable over the
rational numbers.
6x2 + 10x + 5
6(5) = 30
1 30
–1
2 15
–2
3 10
–3
5 6
–5
– 30
–15
–10
–6
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Prime Polynomials
None of the factors of ac add up to b, so this quadratic is
not factorable over the rational numbers. This is a prime
polynomial.
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