Factoring Special Cases

9.2 Learning Objectives

Special Cases - Squares
- Factor a polynomial of the form: ${a}^{2}+2ab+{b}^{2}$
- Factor a polynomial of the form: ${a}^{2}-{b}^{2}$
Special Cases - Cubes
- Factor the sum of cubes.
- Factor the difference of cubes
More Factoring Methods
- Factor expressions with negative exponents
- Factor expressions with fractional exponents
- Factor by substitution
- Factor completely

Why learn how to factor special cases?

Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.

Some people like to find patterns in the world around them, like a game. There are some polynomials that, when factored, follow a specific pattern.

These include:

Perfect square trinomials of the form: ${a}^{2}+2ab+{b}^{2}$

A difference of squares: ${a}^{2}-{b}^{2}$

A sum of cubes: ${a}^{3}+{b}^{3}$

A difference of cubes: ${a}^{3}-{b}^{3}$

In this lesson you will see you can factor each of these types of polynomials following a specific pattern. You will also learn how to factor polynomials that have negative or fractional exponents. Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.

Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.

Some people find it helpful to know when they can take a shortcut to avoid doing extra work. There are some polynomials that will always factor a certain way, and for those we offer a shortcut. Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. The most important skill you will use in this section will be recognizing when you can use the shortcuts.

9.2.1 Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}

We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}

In the following example we will show you how to define a, and b so you can use the shortcut.

Exercises 9.2.A

Factor

25{x}^{2}+20x+4

Answer: First, notice that $25{x}^{2}$ and $4$ are perfect squares because $25{x}^{2}={\left(5x\right)}^{2}$ and $4={2}^{2}$ . This means that $a=5x\text{ and }b=2$ Next, check to see if the middle term is equal to $2ab$ , which it is:

$2ab = 2\left(5x\right)\left(2\right)=20x$ .

Therefore, the trinomial is a perfect square trinomial and can be written as

{\left(a+b\right)}^{2}={\left(5x+2\right)}^{2}

Answer

25{x}^{2}+20x+4={\left(5x+2\right)}^{2}

In the next example, we will show that we can use

1 = 1^2

to factor a polynomial with a term equal to 1.

Example 9.2.B

Factor

49{x}^{2}-14x+1

Answer: First, notice that $49{x}^{2}$ and $1$ are perfect squares because $49{x}^{2}={\left(7x\right)}^{2}$ and $1={1}^{2}$ . This means that $a=7x$ , we could say that $b=1$ , but would that give a middle term of $-14x$ ? We will need to choose $b = -1$ to get the results we want:

$2ab = 2\left(7x\right)\left(-1\right)=-14x$ .

Therefore, the trinomial is a perfect square trinomial and can be written as

{\left(a-b\right)}^{2}={\left(7x-1\right)}^{2}

Answer

49{x}^{2}-14x+1={\left(7x-1\right)}^{2}

In the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula. https://youtu.be/UMCVGDTxxTI We can summarize our process in the following way:

Given a perfect square trinomial, factor it into the square of a binomial.

Confirm that the first and last term are perfect squares.
Confirm that the middle term is twice the product of $ab$ .
Write the factored form as ${\left(a+b\right)}^{2}$ , or ${\left(a-b\right)}^{2}$ .

9.2.2 Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)

We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)

Example 9.2.C

Factor

9{x}^{2}-25

Answer: Notice that $9{x}^{2}$ and $25$ are perfect squares because $9{x}^{2}={\left(3x\right)}^{2}$ and $25={5}^{2}$ . This means that $a=3x,\text{ and }b=5$ The polynomial represents a difference of squares and can be rewritten as $\left(3x+5\right)\left(3x - 5\right)$ . Check that you are correct by multiplying. $\left(3x+5\right)\left(3x - 5\right)=9x^2-15x+15x-25=9x^2-25$

Answer

9{x}^{2}-25=\left(3x+5\right)\left(3x - 5\right)

The most helpful thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.

Example 9.2.D

Factor

81{y}^{2}-144

Answer: Notice that $81{y}^{2}$ and $144$ are perfect squares because $81{y}^{2}={\left(9x\right)}^{2}$ and $144={12}^{2}$ . This means that $a=9x,\text{ and }b=12$ The polynomial represents a difference of squares and can be rewritten as $\left(9x+12\right)\left(9x - 12\right)$ . Check that you are correct by multiplying. $\left(9x+12\right)\left(9x - 12\right)=81x^2-108x+108x-144=81x^2-144$

Answer

81{y}^{2}-144=\left(9x+12\right)\left(9x - 12\right)

In the following video we show another example of how to use the formula for fact a difference of squares. https://youtu.be/Li9IBp5HrFA We can summarize the process for factoring a difference of squares with the shortcut this way:

How To: Given a difference of squares, factor it into binomials.

Confirm that the first and last term are perfect squares.
Write the factored form as $\left(a+b\right)\left(a-b\right)$ .

Think About It

Is there a formula to factor the sum of squares,

a^2+b^2

, into a product of two binomials? Write down some ideas for how you would answer this in the box below before you look at the answer. [practice-area rows="1"][/practice-area]

Answer: There is no way to factor a sum of squares into a product of two binomials, this is because of addition - the middle term needs to "disappear" and the only way to do that is with opposite signs. to get a positive result, you must multiply two numbers with the same signs. The only time a sum of squares can be factored is if they share any common factors, as in the following case: $9x^2+36$ $9x^2={(3x)}^2, \text{ and }36 = 6^2$ The only way to factor this expression is by pulling out the GCF which is 9. $9x^2+36=9(x^2+4)$

9.2.3 Cubes

Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider:

a^{3}+b^{3}

and

a^{3}-b^{3}

. Let’s take a look at how to factor sums and differences of cubes.

Sum of Cubes

The term “cubed” is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width x can be represented by

x^{3}

. (Notice the exponent!) Cubed numbers get large very quickly.

1^{3}=1

2^{3}=8

3^{3}=27

4^{3}=64

, and

5^{3}=125

. Before looking at factoring a sum of two cubes, let’s look at the possible factors. It turns out that

a^{3}+b^{3}

can actually be factored as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

. Let’s check these factors by multiplying.

Example 9.2.E

Does

(a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}

Answer: Apply the distributive property.

$\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)$

Multiply by a.

$\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)$

Multiply by b.

$\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)$

Rearrange terms in order to combine the like terms.

$a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}$

Simplify.

Answer

$a^{3}+b^{3}$

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial

a^{3}+b^{3}

. So, the factors are correct. You can use this pattern to factor binomials in the form

a^{3}+b^{3}

, otherwise known as “the sum of cubes.”

The Sum of Cubes

A binomial in the form

a^{3}+b^{3}

can be factored as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

Examples:

The factored form of

x^{3}+64

\left(x+4\right)\left(x^{2}–4x+16\right)

. The factored form of

8x^{3}+y^{3}

\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)

Example 9.2.F

Factor

x^{3}+8y^{3}

Answer: Identify that this binomial fits the sum of cubes pattern: $a^{3}+b^{3}$ . $a=x$ , and $b=2y$ (since $2y\cdot2y\cdot2y=8y^{3}$ ).

$x^{3}+8y^{3}$

Factor the binomial as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

, substituting

a=x

and

b=2y

into the expression.

$\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)$

Square

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

$\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)$

Multiply

−x\left(2y\right)=−2xy

(writing the coefficient first).

Answer

$\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)$

And that’s it. The binomial

x^{3}+8y^{3}

can be factored as

\left(x+2y\right)\left(x^{2}–2xy+4y^{2}\right)

! Let’s try another one. You should always look for a common factor before you follow any of the patterns for factoring.

Example 9.2.G

Factor

16m^{3}+54n^{3}

Answer: Factor out the common factor 2.

$16m^{3}+54n^{3}$

8m^{3}

and

27n^{3}

are cubes, so you can factor

8m^{3}+27n^{3}

as the sum of two cubes:

a=2m

, and

b=3n

$2\left(8m^{3}+27^n{3}\right)$

Factor the binomial

8m^{3}+27n^{3}

substituting

a=2m

and

b=3n

into the expression

\left(a+b\right)\left(a^{2}-ab+b^{2}\right)

$2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right]$

Square:

(2m)^{2}=4m^{2}

and

(3n)^{2}=9n^{2}

$2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right]$

Multiply

-\left(2m\right)\left(3n\right)=-6mn

Answer

$2\left(2m+3n\right)\left(4m^{2}-6mn+9n^{2}\right)$

Difference of Cubes

Having seen how binomials in the form

a^{3}+b^{3}

can be factored, it should not come as a surprise that binomials in the form

a^{3}-b^{3}

can be factored in a similar way.

The Difference of Cubes

A binomial in the form

a^{3}–b^{3}

can be factored as

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

Examples

The factored form of

x^{3}–64

\left(x–4\right)\left(x^{2}+4x+16\right)

. The factored form of

27x^{3}–8y^{3}

\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)

Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the

+

and

–

signs. Take a moment to compare the factored form of

a^{3}+b^{3}

with the factored form of

a^{3}-b^{3}

. Factored form of

a^{3}+b^{3}

\left(a+b\right)\left(a^{2}-ab+b^{2}\right)

Factored form of

a^{3}-b^{3}

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

This can be tricky to remember because of the different signs—the factored form of

a^{3}+b^{3}

contains a negative, and the factored form of

a^{3}-b^{3}

contains a positive! Some people remember the different forms like this: “Remember one sequence of variables:

a^{3}b^{3}=\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)

. There are 4 missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always

+

.” Try this for yourself. If the first sign is

+

, as in

a^{3}+b^{3}

, according to this strategy how do you fill in the rest:

\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)

? Does this method help you remember the factored form of

a^{3}+b^{3}

and

a^{3}–b^{3}

? Let’s go ahead and look at a couple of examples. Remember to factor out all common factors first.

Example 9.2.H

Factor

8x^{3}–1,000

Answer: Factor out 8.

$8(x^{3}–125)$

Identify that the binomial fits the pattern

a^{3}-b^{3}:a=x

, and

b=5

(since

5^{3}=125

$8\left(x^{3}–125\right)$

Factor

x^{3}–125

\left(a–b\right)\left(a^{2}+ab+b^{2}\right)

, substituting

a=x

and

b=5

into the expression.

$8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right]$

Square the first and last terms, and rewrite

\left(x\right)\left(5\right)

5x

$8\left(x–5\right)\left(x^{2}+5x+25\right)$

Answer

$8\left(x–5\right)\left(x^{2}+5x+25\right)$

Let’s see what happens if you don’t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out. As you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first. Here is one more example. Note that

r^{9}=\left(r^{3}\right)^{3}

and that

8s^{6}=\left(2s^{2}\right)^{3}

Example 9.2.I

Factor

r^{9}-8s^{6}

Answer: Identify this binomial as the difference of two cubes. As shown above, it is. Using the laws of exponents, rewrite $r^{9}$ as $\left(r^{3}\right)^{3}$ .

$r^{9}-8s^{6}$

Rewrite

r^{9}

\left(r^{3}\right)^{3}

and rewrite

8s^{6}

\left(2s^{2}\right)^{3}

$\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}$

Now the binomial is written in terms of cubed quantities. Thinking of

a^{3}-b^{3}

a=r^{3}

and

b=2s^{2}

. Factor the binomial as

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

, substituting

a=r^{3}

and

b=2s^{2}

into the expression.

$\left(r^{3}-2s^{2}\right)\left[\left(r^{3}\right)^{2}+\left(r^{3}\right)\left(2s^{2}\right)+\left(2s^{2}\right)^{2}\right]$

Multiply and square the terms.

$\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)$

Answer

$\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)$

In the following two video examples we show more binomials that can be factored as a sum or difference of cubes. https://youtu.be/tFSEpOB262M https://youtu.be/J_0ctMrl5_0 You encounter some interesting patterns when factoring. Two special cases—the sum of cubes and the difference of cubes—can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:

A binomial in the form $a^{3}+b^{3}$ can be factored as $\left(a+b\right)\left(a^{2}–ab+b^{2}\right)$
A binomial in the form $a^{3}-b^{3}$ can be factored as $\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$

Always remember to factor out any common factors first.

9.2.3 More Factoring Methods

Expressions with fractional or negative exponents can be factored using the same factoring techniques as those with integer exponents. It is important to remember a couple of things first.

When you multiply two exponentiated terms with the same base, you can add the exponents: $x^{-1}\cdot{x^{-1}}=x^{-1+(-1)}=x^{-2}$
When you add fractions, you need a common denominator: $\frac{1}{2}+\frac{1}{3}=\frac{3}{3}\cdot\frac{1}{2}+\frac{2}{2}\cdot\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$
Polynomials have positive integer exponents - if it has a fractional or negative exponent it is an expression.

First, let's practice finding a GCF that is a negative exponent.

Example 9.2.J

Factor

12y^{-3}-2y^{-2}

Answer: If the exponents in this expression were positive we could determine that the GCF is $2y^2$ , but since we have negative exponents, we will need to use $2y^{-2}$ . Therefore $12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1)$ We can check that we are correct by multiplying: $2y^{-2}(6y^{-1}-1)=12y^{-2+(-1)}-2y^{-2}=12y^{-3}-2y^{-2}$

Answer

12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1)

Now let's factor a trinomial that has negative exponents.

Example 9.2.K

Factor

x^{-2}+5x^{-1}+6

Answer: If the exponents on this trinomial were positive, we could factor this as $(x+2)(x+3)$ . Note that the exponent on the x's in the factored form is 1, in other words $(x+2)=(x^{1}+2)$ . Also note that $-1+(-1) = -2$ , therefore if we factor this trinomial as $(x^{-1}+2)(x^{-1}+3)$ , we will get the correct result if we check by multiplying. $(x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6$

Answer

x^{-2}+5x^{-1}+6=(x^{-1}+2)(x^{-1}+3)

In the next example we will see a difference of squares with negative exponents. We can use the same shortcut as we have before, but be careful with the exponent.

Example 9.2.L

Factor

25x^{-4}-36

Answer: Recall that a difference of squares factors in this way: $a^2-b^2=(a-b)(a+b)$ , and the first thing we did was identify a and b to see whether we could factor this as a difference of squares. Given $25x^{-4}-36$ , we can define $a=5x^{-2},\text{ and }b = 6$ because ${5x^{-2}}^2=25x^{-4},\text{ and }6^2=36$ Therefore the factored form is: $(5x^{-2}-6)(5x^{-2}+6)$

Answer

25x^{-4}-36=(5x^{-2}-6)(5x^{-2}+6)

In the following video examples you will see more examples that are similar to the previous three written examples. https://youtu.be/4w99g0GZOCk

9.2.4 Fractional Exponents

Again, we will first practice finding a GCF that has a fractional exponent.

Example 9.2.M

Factor

x^{\frac{2}{3}}+3x^{\frac{1}{3}}

Answer: First, look for the term with the lowest value exponent. In this case, it is $3x^{\frac{1}{3}}$ . Recall that when you multiply terms with exponents, you add the exponents. To get $\frac{2}{3}$ you would need to add $\frac{1}{3}$ to $\frac{1}{3}$ , so we will need a term whose exponent is $\frac{1}{3}$ . $x^{\frac{1}{3}}\cdot{x^{\frac{1}{3}}}=x^{\frac{2}{3}}$ , therefore: $x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+3)$

Answer

x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+3)

In our next example we will factor a perfect square trinomial that has fractional exponents.

Example 9.2.N

Factor

25x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49

Answer: Recall that a perfect square trinomial of the form $a^2+2ab+b^2$ factors as $(a+b)^2$ The first step in factoring a perfect square trinomial was to identify a and b. To find a, we ask: $(?)^2=25x^{\frac{1}{2}}$ , and recall that $(x^a)^b=x^{a\cdot{b}}$ , therefore we are looking for an exponent for x that when multiplied by 2, will give $\frac{1}{2}$ . You can also think about the fact that the middle term is defined as $2ab$ so a will probably have an exponent of $\frac{1}{4}$ , therefore a choice for a may be $5x^{\frac{1}{4}}$ We can check that this is right by squaring a: ${(5x^{\frac{1}{4}})}^{2}=25x^{2\cdot\frac{1}{4}}=25x^{\frac{1}{2}}$ $b = 7\text{ and }b^2=49$ Now we can check whether $2ab =70x^{\frac{1}{4}}$ $2ab=2\cdot{5x^{\frac{1}{4}}}\cdot7=70x^{\frac{1}{4}}$ Our terms work out, so we can use the shortcut to factor: $25x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49=(5x^{\frac{1}{4}}+7)^2$

In our next video example you will see more examples of how to factor expressions with fractional exponents. https://youtu.be/R6BzjR2O4z8

9.2.5 Factor Using Substitution

We are going to move back to factoring polynomials - our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor, but isn't quite the same. Substitution is a useful tool that can be used to "mask" a term or expression to make algebraic operations easier. You may recall that substitution can be used to solve systems of linear equations, and to check whether a point is a solution to a system of linear equations. for example: To determine whether the ordered pair

\left(5,1\right)

is a solution to the given system of equations.

\begin{array}{l}x+3y=8\hfill \\ 2x - 9=y\hfill \end{array}

We can substitute the ordered pair

\left(5,1\right)

into both equations.

$\begin{array}{ll}\left(5\right)+3\left(1\right)=8\hfill & \hfill \\ \text{ }8=8\hfill & \text{True}\hfill \\ 2\left(5\right)-9=\left(1\right)\hfill & \hfill \\ \text{ }\text{1=1}\hfill & \text{True}\hfill \end{array}$

We replaced the variable with a number and then performed the algebraic operations specified. In the next example we will see how we can use a similar technique to factor a fourth degree polynomial.

Example 9.2.O

Factor

x^4+3x^2+2

Answer: This looks a lot like a trinomial that we know how to factor, $x^2+3x+2=(x+2)(x+1)$ , except for the exponents. If we substitute $u=x^2$ , and recognize that $u^2=(x^2)^2=x^4$ we may be able to factor this beast! Everywhere there is an $x^2$ we will replace it with a u, then factor. $u^2+3u+2=(u+1)(u+2)$ We aren't quite done yet, we want to factor the original polynomial which had x as it's variable, so we need to replace $x^2=u$ now that we are done factoring. $(u+1)(u+2)=(x^2+1)(x^2+2)$

Answer

x^4+3x^2+2=(x^2+1)(x^2+2)

In the following video we show two more examples of how to use substitution to factor a fourth degree polynomial and an expression with fractional exponents. https://youtu.be/QUznZt6yrgI

9.2.6 Factor Completely

Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example we will first find the GCF of a trinomial, and after factoring it out we will be able to factor again so that we end up with a product of a monomial, and two binomials.

Example 9.2.p

Factor completely

6m^2k-3mk-3k

Answer: Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is 3k. Factor 3k from the trinomial: $6m^2k-3mk-3k=3k\left(2m^2-m-1\right)$ We are left with a trinomial that can be factored using your choice of factoring method. We will create a table to find the factors of $2\cdot{-1}=-2$ that sum to $-1$

Factors of $2\cdot-1=-2$	Sum of Factors
$2,-1$	$1$
$-2,1$	$-1$

Our factors are

-2,1

, so we can factor by grouping: Rewrite the middle term with the factors we found with the table:

$\left(2m^2-m-1\right)=2m^2-2m+m-1$

Regroup and find the GCF of each group:

$(2m^2-2m)+(m-1)=2m(m-1)+1(m-1)$

Now factor

(m-1)

from each term:

$2m^2-m-1=(m-1)(2m+1)$

Don't forget the original GCF that we factored out! Our final factored form is:

$6m^2k-3mk-3k=3k (m-1)(2m+1)$

In our last example we shoe that it is important to factor out a GCF if there is one before you being using the techniques shown in this module. https://youtu.be/hMAImz2BuPc

Summary

In this section we used factoring with special cases, and factoring by grouping to factor expressions with negative and fractional exponents. We also returned to factoring polynomials and used the substitution method to factor a 4th degree polynomial. The last topic we covered was what it means to factor completely.

学習ガイド > Intermediate Algebra

Factoring Special Cases

9.2 Learning Objectives

Why learn how to factor special cases?

9.2.1 Factoring a Perfect Square Trinomial

A General Note: Perfect Square Trinomials

Exercises 9.2.A

Answer

Example 9.2.B

Answer

Given a perfect square trinomial, factor it into the square of a binomial.

9.2.2 Factoring a Difference of Squares

A General Note: Differences of Squares

Example 9.2.C

Answer

Example 9.2.D

Answer

How To: Given a difference of squares, factor it into binomials.

Think About It

9.2.3 Cubes

Sum of Cubes

Example 9.2.E

Answer

The Sum of Cubes

Examples:

Example 9.2.F

Answer

Example 9.2.G

Answer

Difference of Cubes

The Difference of Cubes

Examples

Example 9.2.H

Answer

Example 9.2.I

Answer

9.2.3 More Factoring Methods

Example 9.2.J

Answer

Example 9.2.K

Answer

Example 9.2.L

Answer

9.2.4 Fractional Exponents

Example 9.2.M

Answer

Example 9.2.N

9.2.5 Factor Using Substitution

Example 9.2.O

Answer

9.2.6 Factor Completely

Example 9.2.p

Summary

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