## Section7.5Factoring Special Polynomials

¶Certain polynomials have patterns that you can train yourself to recognize. And when they have these patterns, there are formulas you can use to factor them, much more quickly than using the techniques from Section 7.3 and Section 7.4.

### Subsection7.5.1Difference of Squares

If \(b\) is some positive integer, then when you multiply \((x-b)(x+b)\text{:}\)

The \(-bx\) and the \(+bx\) cancel each other out. So this is telling us that

And so if we ever encounter a polynomial of the form \(x^2-b^2\) (a “difference of squares”) then we have a quick formula for factoring it. Just identify what “\(b\)” is, and use that in \((x-b)(x+b)\text{.}\)

To use this formula, it's important to recognize which numbers are perfect squares, as in Table 1.3.7.

###### Example7.5.2

Factor \(x^2-16\text{.}\)

The “\(16\)” being subtracted here is a perfect square. It is the same as \(4^2\text{.}\) So we can take \(b=4\) and write:

###### Checkpoint7.5.3

Try to factor one yourself:

We can do a little better. There is nothing special about starting with “\(x^2\)” in these examples. In full generality:

###### Fact7.5.4The Difference of Squares Formula

If \(a\) and \(b\) are any mathematical expressions, then:

###### Example7.5.5

Factor \(1-p^2\text{.}\)

The “\(1\)” at the beginning of this expression is a perfect square; it's the same as \(1^2\text{.}\) The “\(p^2\)” being subtracted here is also perfect square. We can take \(a=1\) and \(b=p\text{,}\) and use The Difference of Squares Formula:

###### Example7.5.6

Factor \(m^2n^2-4\text{.}\)

Is the “\(m^2n^2\)” at the beginning of this expression a perfect square? By the rules for exponents, it is the same as \((mn)^2\text{,}\) so yes, it is a perfect square and we may take \(a=mn\text{.}\) The “\(4\)” being subtracted here is also perfect square. We can take \(b=2\text{.}\) The Difference of Squares Formula tells us:

###### Checkpoint7.5.7

Try to factor one yourself:

###### Example7.5.8

Factor \(x^6-9\text{.}\)

Is the “\(x^6\)” at the beginning of this expression is a perfect square? It may appear to be a *sixth* power, but it is *also* a perfect square because we can write \(x^6=\left(x^3\right)^2\text{.}\) So we may take \(a=x^3\text{.}\) The “\(9\)” being subtracted here is also perfect square. We can take \(b=3\text{.}\) The Difference of Squares Formula tells us:

###### Warning7.5.9

It's a common mistake to write something like \(x^2+16=(x+4)(x-4)\text{.}\) This is not what The Difference of Squares Formula allows you to do, and this is in fact incorrect. The issue is that \(x^2+16\) is a *sum* of squares, not a *difference*. And it happens that \(x^2+16\) is actually prime. In fact, any sum of squares without a common factor will always be prime.

### Subsection7.5.2Perfect Square Trinomials

If we expand \((a+b)^2\text{:}\)

The \(ba\) and the \(ab\) equal each other and double up when added together. So this is telling us that

And so if we ever encounter a polynomial of the form \(a^2+2ab+b^2\) (a “perfect square trinomial”) then we have a quick formula for factoring it.

The tricky part is recognizing when a trinomial you have encountered is in this special form. Ask yourself:

Is the first term a perfect square? If so, jot down what \(a\) would be.

Is the second term a perfect square? If so, jot down what \(b\) would be.

When you multiply \(2\) with what you wrote down for \(a\) and \(b\text{,}\) i.e. \(2ab\text{,}\) do you have the middle term? If you have this middle term exactly, then your polynomial factors as \((a+b)^2\text{.}\) If the middle term is the negative of \(2ab\text{,}\) then the sign on your \(b\) can be reversed, and your polynomial factors as \((a-b)^2\text{.}\)

###### Fact7.5.10The Perfect Square Trinomial Formula

If \(a\) and \(b\) are any mathematical expressions, then:

and

###### Example7.5.11

Factor \(x^2+6x+9\text{.}\)

The first term, \(x^2\text{,}\) is clearly a perfect square. So we could take \(a=x\text{.}\) The last term, \(9\text{,}\) is also a perfect square since it is equal to \(3^2\text{.}\) So we could take \(b=3\text{.}\) Now we multiply \(2ab=2\cdot x\cdot3\text{,}\) and the result is \(6x\text{.}\) This is the middle term, which is what we hope to see.

So we can use The Perfect Square Trinomial Formula:

###### Example7.5.12

Factor \(4x^2-20xy+25y^2\text{.}\)

The first term, \(4x^2\text{,}\) is a perfect square because it equals \((2x)^2\text{.}\) So we could take \(a=2x\text{.}\) The last term, \(25y^2\text{,}\) is also a perfect square since it is equal to \((5y)^2\text{.}\) So we could take \(b=5y\text{.}\) Now we multiply \(2ab=2\cdot (2x)\cdot(5y)\text{,}\) and the result is \(20xy\text{.}\) This is the *negative* of the middle term, which we can work with. The factored form will be \((a-b)^2\) instead of \((a+b)^2\text{.}\)

So we can use The Perfect Square Trinomial Formula:

###### Checkpoint7.5.13

Try to factor one yourself:

###### Warning7.5.14

It is not enough to just see that the first and last terms are perfect squares. For example, \(9x^2+10x+25\) has its first term equal to \((3x)^2\) and its last term equal to \(5^2\text{.}\) But when you examine \(2\cdot(3x)\cdot5\) the result is \(30x\text{,}\) *not* equal to the middle term. So The Perfect Square Trinomial Formula doesn't apply here. In fact, this polynomial doesn't factor at all.

###### Remark7.5.15

To factor these perfect square trinomials, we *could* use methods from Section 7.3 and Section 7.4. As an exercise for yourself, try to factor each of the three previous examples using those methods. The advantage to using The Perfect Square Trinomial Formula is that it is much faster. With some practice, all of the work for using it can be done mentally.

### Subsection7.5.3Sum/Difference of Cubes Formulas

The following calculation may seem to come from nowhere at first, but see it through. If we expand \((a-b)\left(a^2+ab+b^2\right)\text{:}\)

This is telling us that

And so if we ever encounter a polynomial of the form \(a^3-b^3\) (a “difference of cubes”) then we have a quick formula for factoring it.

A similar formula exists for factoring a *sum* of cubes, \(a^3+b^3\text{.}\) Here are both formulas, followed by some tips on how to memorize them.

###### Fact7.5.16The Sum/Difference of Cubes Formula

If \(a\) and \(b\) are any mathematical expressions, then:

and

To memorize this, focus on:

The factorization is a binomial times a trinomial.

The sign you see in the sum/difference of cubes is the same sign you use in the binomial.

The three terms in the trinomial are all quadratic in the variables \(a\) and \(b\) in all the possible ways: \(a^2\text{,}\) \(ab\text{,}\) and \(b^2\text{.}\)

In the trinomial, it's

*always*adding \(a^2\) and \(b^2\text{.}\) But the sign on \(ab\) is always the*opposite*of the sign in the sum/difference of cubes.

###### Differences of cubes *are* sums of cubes

Technically a difference of cubes, \(a^3-b^3\text{,}\) is equal to \(a^3+(-b)^3\text{.}\) If that is confusing, then forget about this and continue. If you understand this, then you can treat any difference of cubes as a sum of cubes, \(a^3+b^3\text{,}\) where \(b\) is negative and you only need to memorize the *sum* of cubes formula.

To use these formulas effectively, we need to recognize when numbers are perfect cubes. Perfect cubes become large fast before you can list too many of them. Try to memorize as many of these as you can:

Let's look at a few examples.

###### Example7.5.17

Factor \(x^3-27\text{.}\)

We recognize that \(x^3\) is a perfect cube and \(27=3^3\text{,}\) so we can use The Sum/Difference of Cubes Formula to factor the binomial. Note that since we have a *difference* of cubes, the binomial factor from the formula will *subtract* two terms, and the middle term from the trinomial will be \(+ab\text{.}\)

###### Example7.5.18

Factor \(27m^3+64n^3\text{.}\)

We recognize that \(27m^3=(3m)^3\) and \(64n^3=(4n)^3\) are both perfect cubes, so we can use The Sum/Difference of Cubes Formula to factor the binomial. Note that since we have a *sum* of cubes, the binomial factor from the formula will *add* two terms, and the middle term from the trinomial will be \(-ab\text{.}\)

###### Checkpoint7.5.19

Try to factor one yourself:

### Subsection7.5.4Factoring in Stages

Sometimes factoring a polynomial will take two or more “stages.” For instance you might use one of the special formulas from this section to factor something into two factors, and *then* those factors might be factor even more. When the task is to *factor* a polynomial, the intention is that you *fully* factor it, breaking down the pieces into even smaller pieces when that is possible.

###### Example7.5.20Factor out any greatest common factor

Factor \(12z^3-27z\text{.}\)

The two terms of this polynomial have greatest common factor \(3z\text{,}\) so the first step in factoring should be to factor this out:

Now we have two factors. There is nothing for us to do with \(3z\text{,}\) but we should ask if \(\left(4z^2-9\right)\) can factor further. And in fact, that is a difference of squares. So we can apply The Difference of Squares Formula. The full process would be:

######
Example7.5.21Recognize a *second* special pattern

Factor \(p^4-1\text{.}\)

Since \(p^4\) is the same as \(\left(p^2\right)^2\text{,}\) we have a difference of squares here. We can apply The Difference of Squares Formula:

It doesn't end here. Of the two factors we found, \(\left(p^2+1\right)\) cannot be factored further. But the other one, \(\left(p^2-1\right)\) is *also* a difference of squares. So we should apply The Difference of Squares Formula again:

###### Checkpoint7.5.22

Try to factor one yourself:

The trinomial from The Sum/Difference of Cubes Formula will *never* factor further. However, in some cases, the binomial from that formula will factor further, and you should look for this.

###### Example7.5.23

Factor \(64p^6-729\text{.}\)

We recognize that \(64p^6\) is a perfect cube because it equals \(\left(4p^2\right)^3\text{.}\) And \(729\) is also a perfect cube because \(729=9^3\text{.}\) So we can use The Sum/Difference of Cubes Formula to factor the binomial. Note that since we have a *difference* of cubes, the binomial factor from the formula will *subtract* two terms, and the middle term from the trinomial will be \(+ab\text{.}\)

As noted, the trinomial here can't possible factor further. But the binomial is a difference of squares, so we must continue.

\begin{align*} \amp=(2p-3)(2p+3)\left(16p^4-36p^2+81\right) \end{align*}###### Example7.5.24

Factor \(32x^6y^2-48x^5y+18x^4\text{.}\)

The first step of factoring any polynomial is to factor out the common factor if possible. For this trinomial, the common factor is \(2x^4\text{,}\) so we write

The square numbers \(16\) and \(9\) in \(16x^2y^2-24xy+9\) hint that maybe we could use The Perfect Square Trinomial Formula. Taking \(a=4xy\) and \(b=3\text{,}\) we multiply \(2ab=2\cdot(4xy)\cdot 3\text{.}\) The result is \(24xy\text{,}\) which is the negative of our middle term. So the whole process is:

### SubsectionExercises

###### 1

Factor the given polynomial

\({y^{2}-1}=\)

###### 2

Factor the given polynomial

\({r^{2}-81}=\)

###### 3

Factor the given polynomial

\({16r^{2}-25}=\)

###### 4

Factor the given polynomial

\({144r^{2}-121}=\)

###### 5

Factor the given polynomial

\({t^{2}r^{2}-9}=\)

###### 6

Factor the given polynomial

\({t^{2}y^{2}-64}=\)

###### 7

Factor the given polynomial

\({x^{2}t^{2}-16}=\)

###### 8

Factor the given polynomial

\({36x^{2}r^{2}-1}=\)

###### 9

Factor the given polynomial

\({9-y^{2}}=\)

###### 10

Factor the given polynomial

\({16-y^{2}}=\)

###### 11

Factor the given polynomial

\({81-49r^{2}}=\)

###### 12

Factor the given polynomial

\({4-9r^{2}}=\)

###### 13

Factor the given polynomial

\({r^{4}-121}=\)

###### 14

Factor the given polynomial

\({t^{4}-49}=\)

###### 15

Factor the given polynomial

\({9t^{4}-25}=\)

###### 16

Factor the given polynomial

\({100x^{4}-1}=\)

###### 17

Factor the given polynomial

\({x^{6}-49}=\)

###### 18

Factor the given polynomial

\({y^{12}-9}=\)

###### 19

Factor the given polynomial

\({36x^{4}-y^{4}}=\)

###### 20

Factor the given polynomial

\({49x^{4}-36y^{4}}=\)

###### 21

Factor the given polynomial

\({x^{10}-64y^{14}}=\)

###### 22

Factor the given polynomial

\({x^{12}-100y^{10}}=\)

###### 23

Factor the given polynomial

\({t^{2}+10t+25}=\)

###### 24

Factor the given polynomial

\({t^{2}+2t+1}=\)

###### 25

Factor the given polynomial

\({x^{2}-18x+81}=\)

###### 26

Factor the given polynomial

\({x^{2}-10x+25}=\)

###### 27

Factor the given polynomial

\({144y^{2}+24y+1}=\)

###### 28

Factor the given polynomial

\({81y^{2}+18y+1}=\)

###### 29

Factor the given polynomial

\({25r^{2}-10r+1}=\)

###### 30

Factor the given polynomial

\({144r^{2}-24r+1}=\)

###### 31

Factor the given polynomial

\({81r^{2}x^{2}-18rx+1}=\)

###### 32

Factor the given polynomial

\({49t^{2}x^{2}-14tx+1}=\)

###### 33

Factor the given polynomial

\({t^{2}+12tx+36x^{2}}=\)

###### 34

Factor the given polynomial

\({x^{2}+8xr+16r^{2}}=\)

###### 35

Factor the given polynomial

\({x^{2}-14xr+49r^{2}}=\)

###### 36

Factor the given polynomial

\({y^{2}-4yt+4t^{2}}=\)

###### 37

Factor the given polynomial

\({64y^{2}+48yr+9r^{2}}=\)

###### 38

Factor the given polynomial

\({16r^{2}+40rx+25x^{2}}=\)

###### 39

Factor the given polynomial

\({36r^{2}-60rt+25t^{2}}=\)

###### 40

Factor the given polynomial

\({121r^{2}-66ry+9y^{2}}=\)

###### 41

Factor the given polynomial

\({t^{3}+27}=\)

###### 42

Factor the given polynomial

\({t^{3}+1000}=\)

\(\text{Hint}: 1000=10^3\)

###### 43

Factor the given polynomial

\({x^{3}-216}=\)

\(\text{Hint}: -216=-6^3\)

###### 44

Factor the given polynomial

\({x^{3}-8}=\)

###### 45

Factor the given polynomial

\({512y^{3}+1}=\)

\(\text{Hint: }512 = 8^3\)

###### 46

Factor the given polynomial

\({125y^{3}+1}=\)

###### 47

Factor the given polynomial

\({8r^{3}-1}=\)

###### 48

Factor the given polynomial

\({512r^{3}-1}=\)

\(\text{Hint: }512 = 8^3\)

###### 49

Factor the given polynomial

\({125r^{3}+343}=\)

\(\text{Hint}: 343=7^3\)

###### 50

Factor the given polynomial

\({1000t^{3}+729}=\)

\(\text{Hint}: 1000=10^3\text{, } 729=9^3\)

###### 51

Factor the given polynomial

\({343t^{3}-125}=\)

\(\text{Hint}: 343=7^3\)

###### 52

Factor the given polynomial

\({64x^{3}-729}=\)

\(\text{Hint}: -729=-9^3\)

###### 53

Factor the given polynomial

\({x^{3}y^{3}+27}=\)

###### 54

Factor the given polynomial

\({x^{3}y^{3}+125}=\)

###### 55

Factor the given polynomial

\({x^{3}-216y^{3}}=\)

\(\text{Hint}: -216=-6^3\)

###### 56

Factor the given polynomial

\({x^{3}-343y^{3}}=\)

\(\text{Hint}: -343=-7^3\)

###### 57

Factor the given polynomial

\({16r^{4}-81}=\)

###### 58

Factor the given polynomial

\({r^{4}-16}=\)

###### 59

Factor the given polynomial

\({3t^{2}-108}=\)

###### 60

Factor the given polynomial

\({3t^{2}-147}=\)

###### 61

Factor the given polynomial

\({8x^{3}-32x}=\)

###### 62

Factor the given polynomial

\({2x^{3}-2x}=\)

###### 63

Factor the given polynomial

\({6y^{3}t^{4}-150yt^{2}}=\)

###### 64

Factor the given polynomial

\({2y^{4}x^{4}-72y^{2}x^{2}}=\)

###### 65

Factor the given polynomial

\({18-2r^{2}}=\)

###### 66

Factor the given polynomial

\({150-6r^{2}}=\)

###### 67

Factor the given polynomial

\({8r^{2}+8r+2}=\)

###### 68

Factor the given polynomial

\({72t^{2}+24t+2}=\)

###### 69

Factor the given polynomial

\({28t^{2}y^{2}+28ty+7}=\)

###### 70

Factor the given polynomial

\({90x^{2}y^{2}+60xy+10}=\)

###### 71

Factor the given polynomial

\({63x^{2}-42x+7}=\)

###### 72

Factor the given polynomial

\({75y^{2}-30y+3}=\)

###### 73

Factor the given polynomial

\({144y^{8}+24y^{7}+y^{6}}=\)

###### 74

Factor the given polynomial

\({81y^{10}+18y^{9}+y^{8}}=\)

###### 75

Factor the given polynomial

\({25r^{6}-10r^{5}+r^{4}}=\)

###### 76

Factor the given polynomial

\({144r^{9}-24r^{8}+r^{7}}=\)

###### 77

Factor the given polynomial

\({100t^{10}+40t^{9}+4t^{8}}=\)

###### 78

Factor the given polynomial

\({96t^{9}+48t^{8}+6t^{7}}=\)

###### 79

Factor the given polynomial

\({63x^{7}-42x^{6}+7x^{5}}=\)

###### 80

Factor the given polynomial

\({36x^{10}-36x^{9}+9x^{8}}=\)

###### 81

Factor the given polynomial

\({27y^{4}+1000y}=\)

\(\text{Hint}: 1000=10^3\)

###### 82

Factor the given polynomial

\({729y^{4}+64y}=\)

\(\text{Hint}: 729=9^3\)

###### 83

Factor the given polynomial

\({216x^{3}+125y^{3}}=\)

\(\text{Hint}: 216=6^3\)

###### 84

Factor the given polynomial

\({343x^{3}+27y^{3}}=\)

\(\text{Hint}: 343=7^3\)

###### 85

Factor the given polynomial

\({3r^{4}-243}=\)

###### 86

Factor the given polynomial

\({5t^{4}-80}=\)

###### 87

Factor the given polynomial

\({t^{7}+27t^{4}}=\)

###### 88

Factor the given polynomial

\({x^{10}+1000x^{7}}=\)

\(\text{Hint}: 1000=10^3\)

###### 89

Factor the given polynomial

\({x^{2}+36}=\)

###### 90

Factor the given polynomial

\({y^{2}+4}=\)

###### 91

Factor the given polynomial

\({9y^{3}+9y}=\)

###### 92

Factor the given polynomial

\({10y^{3}+40y}=\)

###### 93

Factor the given polynomial

\({0.01r-r^{3}}=\)

###### 94

Factor the given polynomial

\({0.49r-r^{3}}=\)

###### 95

Factor the given polynomial

\({\left(t-3\right)^{2}-100}=\)

###### 96

Factor the given polynomial

\({\left(t+10\right)^{2}-9}=\)

###### 97

Factor the given polynomial

\({x^{2}-20x+100-81y^{2}}=\)

###### 98

Factor the given polynomial

\({x^{2}-14x+49-25y^{2}}=\)