Cubic Polynomials
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C HAPTER
Chapter 1. Cubic Polynomials
1
Cubic Polynomials
Introduction
In this lesson you will learn to factor cubic polynomials by removing a common factor. You will also learn to factor
these polynomials by grouping. The final concept that you will learn will be to apply factoring by grouping to special
products.
Objectives
The lesson objectives for Factoring Polynomials are:
• Factoring by Removing a Common Factor
• Factoring by Grouping
• Factoring Special Products
Factoring a Common Factor
Introduction
In this concept you will be working with polynomials of the third degree. Polynomials where the largest exponent on
the variable is three (3) are known as cubics. Therefore a cubic polynomial is a polynomial of degree equal to 3. An
example could include 9x3 + 10x − 5. Another example is 8x3 + 2x2 − 5x − 7. In this first concept of lesson Cubic
Polynomials, you will be factoring cubics by removing a common factor. Recall in lesson Factoring Polynomials
when you removed common factors from quadratic (trinomial) functions. The same skills will be applied here.
You use the distributive property to factor out the greatest common factor (GCF) when factoring cubics that have
common terms in the polynomials.
Watch This
Khan Academy Factoring and the Distributive Property
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/59347
Guidance
Factor completely the following polynomial: 8x3 + 24x2 − 32x.
Look for the common factors in each of the terms.
8x3 = 8 · x · x · x
24x2 = 8 · 3 · x · x
−32x = 8 · −4 · x
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Therefore:
8x3 + 24x2 + 32x = 8x(x2 + 3x − 4)
= 8x(x + 4)(x − 1)
Example A
Factor completely the following polynomial: 3x3 − 15x.
Look for the common factors in each of the terms.
3x3 = 3 · x · x · x
−15x = 3 · −5 · x
Therefore:
3x3 − 15x = 3x(x2 − 5)
Example B
Factor completely the following polynomial: 2a3 + 16a2 + 8a.
Look for the common factors in each of the terms.
2a3 = 2 · a · a · a
16a2 = 2 · 8 · a · a
8a = 2 · 4 · a
Therefore:
2a3 + 16a2 + 8a = 2a(a2 + 8a + 4)
Example C
Factor completely the following polynomial: 6s3 + 36s2 − 18s − 42.
Look for the common factors in each of the terms.
6s3 = 6 · s · s · s
36s2 = 6 · 6 · s · s
−18s = 6 · −3 · s
−42 = 6 · −7
Therefore:
6s3 + 36s2 − 18s − 42 = 6(s3 + 6s2 − 3s − 7).
Vocabulary
Cubic Polynomial
A cubic polynomial is a polynomial of degree equal to 3. For example 8x3 +2x2 −5x−7 is a cubic polynomial.
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Chapter 1. Cubic Polynomials
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number
and a sum is equal to the sum of the individual products of the number and the addends.
For example, in the
expression: 23 (x + 5), the distributive property states that the product of a number 32 and a sum (x + 5) is
equal to the sum of the individual products of the number 23 and the addends (x and 5).
Greatest Common Factor
The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly)
each of the terms of the polynomial.
Guided Practice
1. Factor completely the following polynomial: 9w3 + 12w.
2. Factor completely the following polynomial: y3 + 4y2 + 4y.
3. Factor completely the following polynomial: 2t 3 − 10t 2 + 8t.
Answers
1. 9w3 + 12w
Look for the common factors in each of the terms.
9w3 = 3 · 3 · w · w · w
12w = 3 · 4 · w
Therefore:
9w3 + 12w = 3w(3w2 + 4)
2. y3 + 4y2 + 4y
Look for the common factors in each of the terms.
y3 = y · y · y
4y2 = 4 · y · y
4y = 4 · y
Therefore:
y3 + 4y2 + 4y = y(y2 + 4y + 4)
3. 2t 3 − 10t 2 + 8t
Look for the common factors in each of the terms.
2t 3 = 2 · t · t · t
−10t 2 = 2 · −5 · t · t
8t = 2 · 4 · t
Therefore:
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2t 3 − 10t 2 + 8t = 2t(t 2 − 5t + 4)
= 2t(t − 4)(t − 1)
Summary
In this concept you may have noticed that you used similar problem solving strategies to those you used in lesson
Factoring Polynomials. You began your study of cubic polynomials by factoring out the greatest common factor
(the GCF). This strategy you learned earlier.
One of the easiest ways to visualize the greatest common factor is to isolate each term and break each term down
into its smallest parts. In this way, you can quickly see which factors are common to all terms in the polynomial and
therefore factor them out of the cubic.
Problem Set
Factor completely each of the following polynomials:
1.
2.
3.
4.
5.
6x3 − 12
4x3 − 12x2
8y3 + 32y
15a3 + 30a2
21q3 + 63q
Factor completely each of the following polynomials:
1.
2.
3.
4.
5.
4x3 − 12x2 − 8
12e3 + 24e2 − 6
15s3 − 30s + 45
22r3 − 11r2 + 44
32d 3 − 16d 2 + 12d
Factor completely each of the following polynomials:
1.
2.
3.
4.
5.
5x3 + 15x2 + 25x − 30
3y3 − 9y2 + 27y + 36
12s3 − 24s2 + 36s − 48
8x3 + 24x2 + 80x
5x3 − 25x2 − 70x
Factoring by Grouping
Introduction
Another way that you can use the distributive property to factor polynomials is to factor by grouping. Factoring by
grouping is used when you have polynomials with four terms. It cannot be used all of the time but when it can,
finding the greatest common factor is possible. In factoring by grouping you factor the terms into two groups and
then take the greatest common factor out of one of them.
In this concept you will learn to work with finding the greatest common factor of cubic polynomials using the
concept of factoring by grouping. In factoring by grouping, which is different from what you have just learned, you
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Chapter 1. Cubic Polynomials
have to first group the terms into two groups. You then factor each group, use your distributive property, then factor
any quadratics that remain (and are possible). Let’s begin exploring factoring by grouping cubic polynomials!
Watch This
Khan Academy Factoring by Grouping
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/123
Note: The above video shows factoring by grouping of quadratic (trinomial) expressions. The same problem solving
concept will be developed in this lesson for cubic polynomials.
Guidance
A tank is bought at the pet store and is known to have a volume of 12 cubic feet. The dimensions are shown in the
diagram below. If your new pet requires the tank to be at least 3 feet high, did you buy a big enough tank?
To solve this problem, you need to calculate the volume of the tank.
V = l ×w×h
12 = (x + 4) × (x − 1) × (x)
12 = (x2 + 3x − 4) × (x)
12 = x3 + 3x2 − 4x
0 = x3 + 3x2 − 4x − 12
Now you need to factor by grouping.
0 = (x3 + 3x2 ) − (4x + 12)
Factor out the common terms in each of the sets of brackets.
0 = x2 (x + 3) − 4(x + 3)
Factor out the group of terms (x + 3) from the expression.
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0 = (x + 3)(x2 − 4)
Completely factor the remaining quadratic expression.
0 = (x + 3)(x − 2)(x + 2)
Now solve for the variable x.
0 = (x + 3) (x − 2) (x + 2)
.
x+3 = 0
x = −3
↓
x−2 = 0
&
x+2 = 0
x=2
x = −2
Since you are looking for a length, only x = 2 is a good solution. But since you need a tank 3 feet high and this one
is only 2 feet high, you need to go back to the pet shop and buy a bigger one.
Example A
Factor the following polynomial by grouping:
w3 − 2w2 − 9w + 18.
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
w3 − 2w2 − 9w + 18 = (w3 − 2w2 ) − (9w − 18)
Step 2: Factor out the common terms in each of the sets of brackets.
(w3 − 2w2 ) − (9w − 18) = w2 (w − 2) − 9(w − 2)
Step 3: Use the distributive property to factor out the common group (w − 2).
w2 (w − 2) − 9(w − 2) = (w − 2)(w2 − 9)
Step 4: Completely factor the remaining quadratic expression (w2 − 9).
(w − 2)(w2 − 9) = (w − 2)(w + 3)(w − 3)
Therefore:
w3 − 2w2 − 9w + 18 = (w − 2)(w + 3)(w − 3)
Example B
Factor the following polynomial by grouping:
2s3 − 8s2 + 3s − 12
Step 1: Group the terms into two groups.
2s3 − 8s2 + 3s − 12 = (2s3 − 8s2 ) + (3s − 12)
Step 2: Factor out the common terms in each of the sets of brackets.
(2s3 − 8s2 ) + (3s − 12) = 2s2 (s − 4) + 3(s − 4)
Step 3: Use the distributive property to factor out the common group (s − 4).
2s2 (s − 4) + 3(s − 4) = (s − 4)(2s2 + 3)
Step 4: The expression (2s3 + 3) cannot be factored.
Therefore:
2s3 − 8s2 + 3s − 12 = (s − 4)(2s2 + 3)
Example C
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Chapter 1. Cubic Polynomials
Factor the following polynomial by grouping: y3 + 5y2 − 4y − 20.
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
y3 + 5y2 − 4y − 20 = (y3 + 5y2 ) − (4y + 20)
Step 2: Factor out the common terms in each of the sets of brackets.
(y3 + 5y2 ) − (4y + 20) = y2 (y + 5) − 4(y + 5)
Step 3: Use the distributive property to factor out the common group (y + 5).
y2 (y + 5) − 4(y + 5) = (y + 5)(y2 − 4)
Step 4: Completely factor the remaining quadratic expression (y2 − 4).
(y + 5)(y2 − 4) = (y + 5)(y + 2)(y − 2)
Therefore:
y3 + 5y2 − 4y − 20 = (y + 5)(y + y)(y − 2)
Vocabulary
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number
and a sum is equal to the sum of the individual products of the number and the addends.
For example, in the
2
2
expression: 3 (x + 5), the distributive property states that the product of a number 3 and a sum (x + 5) is
equal to the sum of the individual products of the number 23 and the addends (x and 5).
Greatest Common Factor
The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly)
each of the terms of the polynomial.
Guided Practice
1. Factor the following polynomial by grouping: y3 − 4y2 − 4y + 16.
2. Factor the following polynomial by grouping: 3x3 − 4x2 − 3x + 4.
3. Factor the following polynomial by grouping: e3 + 3e − 4e − 12.
Answers
1. y3 − 4y2 − 4y + 16
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
y3 − 4y2 − 4y + 12 = (y3 − 4y2 ) − (4y − 16)
Step 2: Factor out the common terms in each of the sets of brackets.
(y3 − 4y2 ) − (4y − 12) = y2 (y − 4) − 4(y − 4)
Step 3: Use the distributive property to factor out the common group (y − 4).
y2 (y − 4) − 4(y − 4) = (y − 4)(y2 − 4)
Step 4: Completely factor the remaining quadratic expression (y2 − 4).
(y − 4)(y2 − 4) = (y − 4)(y + 2)(y − 2)
Therefore:
y3 − 4y2 − 4y + 16 = (y − 4)(y + 2)(y − 2)
2. 3x3 − 4x2 − 3x + 4
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Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
3x3 − 4x2 − 3x + 4 = (3x3 − 4x2 ) − (3x − 4)
Step 2: Factor out the common terms in each of the sets of brackets.
(3x3 − 4x2 ) − (3x − 4) = x2 (3x − 4) − 1(3x − 4)
Step 3: Use the distributive property to factor out the common group (3x − 4).
x2 (3x − 4) − 1(3x − 4) = (3x − 4)(x2 − 1)
Step 4: Completely factor the remaining quadratic expression (x2 − 1).
(3x − 4)(x2 − 1) = (3x − 4)(x + 1)(x − 1)
Therefore:
3x3 − 4x2 − 3x + 4 = (3x − 4)(x + 1)(x − 1)
3. e3 + 3e − 4e − 12
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
e3 + 3e2 − 4e − 12 = (e3 + 3e2 ) − (4e + 12)
Step 2: Factor out the common terms in each of the sets of brackets.
(e3 + 3e2 ) − (4e + 12) = e2 (e + 3) − 4(e + 3)
Step 3: Use the distributive property to factor out the common group (e + 3).
e2 (e + 3) − 4(e + 3) = (e + 3)(e2 − 4)
Step 4: Completely factor the remaining quadratic expression (e2 − 4).
(e + 3)(e2 − 4) = (e + 3)(e + 2)(e − 2)
Therefore:
e3 + 3e2 − 4e − 12 = (e + 3)(e + 2)(e − 2)
Summary
To factor cubic polynomials by grouping involves four steps, one of which is the distributive property. The distributive property is something you have been learning for a long time in Algebra, and its application in cubic polynomials
is just one more way it shows its usefulness. Factoring cubic polynomials involves problem solving skills that you
have learned in previous lessons such as factoring quadratics, finding greatest common factors, and combining like
terms. For these problems, however, you have the opportunity to combine all of these skills in one problem solving
experience!
To solve cubic polynomials by grouping remember the four steps:
Step 1: Group the terms into two groups. Remember to watch for the sign change when you are grouping the second
set of terms. Placing the negative outside a bracket will change the signs inside the bracket.
Step 2: Factor out the common terms in each of the sets of brackets.
Step 3: Use the distributive property to factor out the common group (the binomial).
Step 4: Completely factor the remaining quadratic expression.
Problem Set
Factor the following cubic polynomials by grouping:
1. x3 − 3x2 − 36x + 108
2. e3 − 3e2 − 81e + 243
3. x3 − 10x2 − 49x + 490
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Chapter 1. Cubic Polynomials
4. y3 − 7y2 − 5y + 35
5. x3 + 9x2 + 3x + 27
Factor the following cubic polynomials by grouping:
1.
2.
3.
4.
5.
3x3 + x2 − 3x − 1
5s3 − 6s2 − 45s + 54
4a3 − 7a2 + 4a − 7
5y2 + 15y2 − 45y − 135
3x3 + 15x2 − 12x − 60
Factor the following cubic polynomials by grouping:
1.
2.
3.
4.
5.
2e3 + 14e2 + 7e + 49
2k3 + 16k2 + 38k + 24
−6x3 + 3x2 + 54x − 27
−5m3 − 6m2 + 20m + 24
−2x3 − 8x2 + 14x + 56
Factoring Special Products
Introduction
Just as you learned the special products for quadratic (trinomial) expressions, there are special products for cubic
polynomials. Remember there were three special products for quadratic (trinomial) expressions. For cubic polynomials, there are two special products. The first is the sum of two cubes. The second is the difference of two cubes.
These two special products are shown in the box below.
The sum of two cubes
x3 + y3 = (x + y)(x2 − xy + y2 )
The difference of two cubes
x3 − y3 = (x − y)(x2 + xy + y2 )
In this concept you will learn to factor the two types of special products in cubic polynomials. As it was with the
quadratic expressions, being able to recognize that the cubic polynomial is indeed a special product means that you
can quickly factor the polynomial. Take a moment to commit these two special cases to memory so that solving
these types of cubic polynomials will be more efficient.
Guidance
Factor the following cubic polynomial: 375x3 + 648.
To solve this problem you need to first recognize that this is a special case cubic polynomial, namely x3 + y3 =
(x + y)(x2 − xy + y2 ).
Next if you look at this polynomial, you see that first you can factor out the greatest common factor.
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375x3 + 648 :
375x3 = 3 · 125x3
648 = 3 · 216
Therefore 375x3 + 648 = 3(125x3 + 216)
Now let’s factor the special case polynomial. Remember that you have a number (numerical coefficient) attached to
the x variable. This means that you have to include it in the factoring. In other words for now becomes (x + y) =
(ax + y), and (x2 − xy + y2 ) now becomes (a2 x2 − axy + y2 ).
375x3 + 648 = 3(125x3 + 216)
375x3 + 648 = 3(5x + 6)(25x2 − 30x + 36)
Example A
x3 + 27
This is the sum of two cubes or x3 + y3 = (x + y)(x2 − xy + y2 ).
Step 1: Put the cubic polynomial into the sum of two cubes form:
x3 + 27 = x3 + y3
Step 2: Factor out the sum of two cubes
x3 + 33 = (x + 3)(x2 − 3x + 9).
Example B
x3 − 343
This is the difference of two cubes or x3 − y3 = (x − y)(x2 + xy + y2 ).
Step 1: Put the cubic polynomial into the sum of two cubes form:
x3 − 343 = x3 − 73
Step 2: Factor out the sum of two cubes
x3 − 343 = (x − 7)(x2 + 7x + 49).
Example C
64x3 − 1
This is the difference of two cubes or x3 − y3 = (x − y)(x2 + xy + y2 ).
Step 1: Put the cubic polynomial into the sum of two cubes form:
64x3 − 1 = 43 x3 − 13
Step 2: Factor out the sum of two cubes
64x3 − 1 = (4x − 1)(42 x2 + 4x + 12 ).
64x3 − 1 = (4x − 1)(16x2 + 4x + 1).
Vocabulary
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number
and a sum is equal to the sum of the individual products of the number and the addends. For example, in
the expression: 3(x + 5), the distributive property states that the product of a number (3) and a sum (x + 5) is
equal to the sum of the individual products of the number (3) and the addends (x and 5).
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Chapter 1. Cubic Polynomials
Difference of Two Cubes
The difference of two cubes is a special product cubic polynomial in the form of x3 − y3 . This type of
polynomial can be quickly factored using the expression:
(x3 − y3 ) = (x − y)(x2 + xy + y2 )
Sum of Two Cubes
The sum of two cubes is a special product cubic polynomial in the form of x3 + y3 . This type of polynomial
can be quickly factored using the expression:
(x3 + y3 ) = (x + y)(x2 − xy + y2 )
Guided Practice
1. Factor the following special product cubic polynomial: x3 + 512
2. Factor the following special product cubic polynomial: 8x3 + 125
3. Factor the following special product cubic polynomial: x3 − 216
Answers
1. x3 + 512
This is the sum of two cubes or x3 + y3 = (x + y)(x2 − xy + y2 ).
Step 1: Put the cubic polynomial into the sum of two cubes form: x3 + 512 = x3 + 83
Step 2: Factor out the sum of two cubes x3 + 83 = (x + 8)(x2 − 8x + 64).
2. 8x3 + 125
This is the sum of two cubes or x3 + y3 = (x + y)(x2 − xy + y2 )
Step 1: Put the cubic polynomial into the sum of two cubes form: 8x3 + 125 = 23 x3 + 53
Step 2: Factor out the sum of two cubes 23 x3 + 53 = (2x + 5)(4x2 − 10x + 25).
3. x3 − 216
This is the difference of two cubes or x3 − y3 = (x − y)(x2 + xy + y2 ).
Step 1: Put the cubic polynomial into the sum of two cubes form: x3 − 216 = x3 − 63
Step 2: Factor out the sum of two cubes x3 − 63 = (x − 6)(x2 + 6x + 36).
Summary
In lesson Operations with Polynomials you learned about special cases in quadratic expressions or polynomials
with a degree of two. In this final concept of lesson Cubic Polynomials, you learned about the special cases
for polynomials with a degree of 3. The box below summarizes the special cases for both quadratic and cubic
polynomials.
Special Case 1: (x + y)2 = x2 + 2xy + y2
Example: (x + 5)2 = x2 + 10x + 25
Special Case 2: (x − y)2 = x2 − 2xy + y2
Example: (2x − 8)2 = 4x2 − 32x + 64
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Special Case 3: (x + y)(x − y) = x2 − y2
Example: (5x + 10)(5x − 10) = 25x2 − 100
Sum of two cubes: (x3 + y3 ) = (x + y)(x2 − xy + y2 )
Example: (x3 + 125) = (x + 5)(x2 − 5x + 25)
Difference of two cubes: (x3 − y3 ) = (x − y)(x2 + xy + y2 )
Example: (x3 − 125) = (x − 5)(x2 + 5x + 25)
Learning these special cases and being able to recognize them allows you to quickly factor such polynomials in
problem solving situations.
Problem Set
Factor the following special product cubic polynomials:
1.
2.
3.
4.
5.
x3 + h3
a3 + 125
8x3 + 64
x3 + 1728
2x3 + 6750
Factor the following special product cubic polynomials:
1.
2.
3.
4.
5.
h3 − 64
s3 − 216
p3 − 512
4e3 − 32
2w3 − 250
Factor the following special product cubic polynomials:
1.
2.
3.
4.
5.
x3 + 8
y3 − 1
125e3 − 8
64a3 + 2197
54z3 + 3456
Summary
In this lesson, you have worked with cubic polynomials. Cubic polynomials are polynomials of degree three.
Examples include x3 + 8, x3 − 4x2 + 3x − 5, and so on. Notice in these examples, the largest exponent for the variable
is three (3). They are all cubics. You worked with three concepts in this lesson. The first concept involved factoring
a common factor. You worked on this problem solving strategy earlier with quadratic (trinomial) expressions, and
the same procedure is involved with cubic polynomials.
The second concept required you to factor by grouping. This concept can be done with quadratics as well as cubics.
It allows you to quickly factor a cubic by grouping the four terms into two groups and then factoring out a common
binomial from each group.
The last concept introduced you to the two special products of cubic polynomials. You learned three special cases
for quadratics and now you have two special cases for cubics. What is beneficial about these special cases is that
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Chapter 1. Cubic Polynomials
they allow you to quickly factor the cubic polynomials by following the associated rule. Knowing how to identify
polynomials in these special cases will allow you to quickly factor both quadratic and cubic polynomials.
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