Factoring Special Cases
9.2 Learning Objectives
- Special Cases - Squares
- Factor a polynomial of the form: [latex]{a}^{2}+2ab+{b}^{2}[/latex]
- Factor a polynomial of the form: [latex]{a}^{2}-{b}^{2}[/latex]
- 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?
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: [latex]{a}^{2}+2ab+{b}^{2}[/latex]
A difference of squares: [latex]{a}^{2}-{b}^{2}[/latex]
A sum of cubes: [latex]{a}^{3}+{b}^{3}[/latex]
A difference of cubes: [latex]{a}^{3}-{b}^{3}[/latex]
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.A General Note: Perfect Square Trinomials
A perfect square trinomial can be written as the square of a binomial:Exercises 9.2.A
Factor [latex]25{x}^{2}+20x+4[/latex].Answer: First, notice that [latex]25{x}^{2}[/latex] and [latex]4[/latex] are perfect squares because [latex]25{x}^{2}={\left(5x\right)}^{2}[/latex] and [latex]4={2}^{2}[/latex]. This means that [latex]a=5x\text{ and }b=2[/latex] Next, check to see if the middle term is equal to [latex]2ab[/latex], which it is:
[latex]2ab = 2\left(5x\right)\left(2\right)=20x[/latex].
Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a+b\right)}^{2}={\left(5x+2\right)}^{2}[/latex].Answer
[latex-display]25{x}^{2}+20x+4={\left(5x+2\right)}^{2}[/latex-display]Example 9.2.B
Factor [latex]49{x}^{2}-14x+1[/latex].Answer: First, notice that [latex]49{x}^{2}[/latex] and [latex]1[/latex] are perfect squares because [latex]49{x}^{2}={\left(7x\right)}^{2}[/latex] and [latex]1={1}^{2}[/latex]. This means that [latex]a=7x[/latex], we could say that [latex]b=1[/latex], but would that give a middle term of [latex]-14x[/latex]? We will need to choose [latex]b = -1[/latex] to get the results we want:
[latex]2ab = 2\left(7x\right)\left(-1\right)=-14x[/latex].
Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a-b\right)}^{2}={\left(7x-1\right)}^{2}[/latex].Answer
[latex-display]49{x}^{2}-14x+1={\left(7x-1\right)}^{2}[/latex-display]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 [latex]ab[/latex].
- Write the factored form as [latex]{\left(a+b\right)}^{2}[/latex], or[latex]{\left(a-b\right)}^{2}[/latex].
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 General Note: Differences of Squares
A difference of squares can be rewritten as two factors containing the same terms but opposite signs.Example 9.2.C
Factor [latex]9{x}^{2}-25[/latex].Answer: Notice that [latex]9{x}^{2}[/latex] and [latex]25[/latex] are perfect squares because [latex]9{x}^{2}={\left(3x\right)}^{2}[/latex] and [latex]25={5}^{2}[/latex]. This means that [latex]a=3x,\text{ and }b=5[/latex] The polynomial represents a difference of squares and can be rewritten as [latex]\left(3x+5\right)\left(3x - 5\right)[/latex]. Check that you are correct by multiplying. [latex-display]\left(3x+5\right)\left(3x - 5\right)=9x^2-15x+15x-25=9x^2-25[/latex-display]
Answer
[latex-display]9{x}^{2}-25=\left(3x+5\right)\left(3x - 5\right)[/latex-display]Example 9.2.D
Factor [latex]81{y}^{2}-144[/latex].Answer: Notice that [latex]81{y}^{2}[/latex] and [latex]144[/latex] are perfect squares because [latex]81{y}^{2}={\left(9x\right)}^{2}[/latex] and [latex]144={12}^{2}[/latex]. This means that [latex]a=9x,\text{ and }b=12[/latex] The polynomial represents a difference of squares and can be rewritten as [latex]\left(9x+12\right)\left(9x - 12\right)[/latex]. Check that you are correct by multiplying. [latex-display]\left(9x+12\right)\left(9x - 12\right)=81x^2-108x+108x-144=81x^2-144[/latex-display]
Answer
[latex-display]81{y}^{2}-144=\left(9x+12\right)\left(9x - 12\right)[/latex-display]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 [latex]\left(a+b\right)\left(a-b\right)[/latex].
Think About It
Is there a formula to factor the sum of squares, [latex]a^2+b^2[/latex], 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: [latex-display]9x^2+36[/latex-display] [latex-display]9x^2={(3x)}^2, \text{ and }36 = 6^2[/latex-display] The only way to factor this expression is by pulling out the GCF which is 9. [latex-display]9x^2+36=9(x^2+4)[/latex-display]
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: [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}-b^{3}[/latex]. 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 [latex]x^{3}[/latex]. (Notice the exponent!) Cubed numbers get large very quickly. [latex]1^{3}=1[/latex], [latex]2^{3}=8[/latex], [latex]3^{3}=27[/latex], [latex]4^{3}=64[/latex], and [latex]5^{3}=125[/latex]. Before looking at factoring a sum of two cubes, let’s look at the possible factors. It turns out that [latex]a^{3}+b^{3}[/latex] can actually be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]. Let’s check these factors by multiplying.Example 9.2.E
Does [latex](a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}[/latex]?Answer: Apply the distributive property.
[latex]\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)[/latex]
Multiply by a.[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)[/latex]
Multiply by b.[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)[/latex]
Rearrange terms in order to combine the like terms.[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[/latex]
Simplify.Answer
[latex]a^{3}+b^{3}[/latex]
The Sum of Cubes
A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex].Examples:
The factored form of [latex]x^{3}+64[/latex] is [latex]\left(x+4\right)\left(x^{2}–4x+16\right)[/latex]. The factored form of [latex]8x^{3}+y^{3}[/latex] is [latex]\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)[/latex].Example 9.2.F
Factor [latex]x^{3}+8y^{3}[/latex].Answer: Identify that this binomial fits the sum of cubes pattern: [latex]a^{3}+b^{3}[/latex]. [latex]a=x[/latex], and [latex]b=2y[/latex] (since [latex]2y\cdot2y\cdot2y=8y^{3}[/latex]).
[latex]x^{3}+8y^{3}[/latex]
Factor the binomial as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=2y[/latex] into the expression.[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)[/latex]
Square [latex](2y)^{2}=4y^{2}[/latex].[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)[/latex]
Multiply [latex]−x\left(2y\right)=−2xy[/latex] (writing the coefficient first).Answer
[latex]\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)[/latex]
Example 9.2.G
Factor [latex]16m^{3}+54n^{3}[/latex].Answer: Factor out the common factor 2.
[latex]16m^{3}+54n^{3}[/latex]
[latex]8m^{3}[/latex] and [latex]27n^{3}[/latex] are cubes, so you can factor [latex]8m^{3}+27n^{3}[/latex] as the sum of two cubes: [latex]a=2m[/latex], and [latex]b=3n[/latex].[latex]2\left(8m^{3}+27^n{3}\right)[/latex]
Factor the binomial [latex]8m^{3}+27n^{3}[/latex] substituting [latex]a=2m[/latex] and [latex]b=3n[/latex] into the expression [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex].[latex]2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right][/latex]
Square: [latex](2m)^{2}=4m^{2}[/latex] and [latex](3n)^{2}=9n^{2}[/latex].[latex]2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right][/latex]
Multiply [latex]-\left(2m\right)\left(3n\right)=-6mn[/latex].Answer
[latex]2\left(2m+3n\right)\left(4m^{2}-6mn+9n^{2}\right)[/latex]
Difference of Cubes
Having seen how binomials in the form [latex]a^{3}+b^{3}[/latex] can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[/latex] can be factored in a similar way.The Difference of Cubes
A binomial in the form [latex]a^{3}–b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex].Examples
The factored form of [latex]x^{3}–64[/latex] is [latex]\left(x–4\right)\left(x^{2}+4x+16\right)[/latex]. The factored form of [latex]27x^{3}–8y^{3}[/latex] is [latex]\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)[/latex].Example 9.2.H
Factor [latex]8x^{3}–1,000[/latex].Answer: Factor out 8.
[latex]8(x^{3}–125)[/latex]
Identify that the binomial fits the pattern [latex]a^{3}-b^{3}:a=x[/latex], and [latex]b=5[/latex] (since [latex]5^{3}=125[/latex]).[latex]8\left(x^{3}–125\right)[/latex]
Factor [latex]x^{3}–125[/latex] as [latex]\left(a–b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=5[/latex] into the expression.[latex]8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right][/latex]
Square the first and last terms, and rewrite [latex]\left(x\right)\left(5\right)[/latex] as [latex]5x[/latex].[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]
Answer
[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]
Example 9.2.I
Factor [latex]r^{9}-8s^{6}[/latex].Answer: Identify this binomial as the difference of two cubes. As shown above, it is. Using the laws of exponents, rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex].
[latex]r^{9}-8s^{6}[/latex]
Rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex] and rewrite [latex]8s^{6}[/latex] as [latex]\left(2s^{2}\right)^{3}[/latex].[latex]\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}[/latex]
Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[/latex], [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex]. Factor the binomial as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex] into the expression.[latex]\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][/latex]
Multiply and square the terms.[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]
Answer
[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]
- A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]
- A binomial in the form [latex]a^{3}-b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex]
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: [latex]x^{-1}\cdot{x^{-1}}=x^{-1+(-1)}=x^{-2}[/latex]
- When you add fractions, you need a common denominator: [latex]\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}[/latex]
- Polynomials have positive integer exponents - if it has a fractional or negative exponent it is an expression.
Example 9.2.J
Factor [latex]12y^{-3}-2y^{-2}[/latex]Answer: If the exponents in this expression were positive we could determine that the GCF is [latex]2y^2[/latex], but since we have negative exponents, we will need to use [latex]2y^{-2}[/latex]. Therefore [latex]12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1)[/latex] We can check that we are correct by multiplying: [latex-display]2y^{-2}(6y^{-1}-1)=12y^{-2+(-1)}-2y^{-2}=12y^{-3}-2y^{-2}[/latex-display]
Answer
[latex-display]12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1)[/latex-display]Example 9.2.K
Factor [latex]x^{-2}+5x^{-1}+6[/latex].Answer: If the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[/latex]. Note that the exponent on the x's in the factored form is 1, in other words [latex](x+2)=(x^{1}+2)[/latex]. Also note that [latex]-1+(-1) = -2[/latex], therefore if we factor this trinomial as [latex](x^{-1}+2)(x^{-1}+3)[/latex], we will get the correct result if we check by multiplying. [latex-display](x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6[/latex-display]
Answer
[latex-display]x^{-2}+5x^{-1}+6=(x^{-1}+2)(x^{-1}+3)[/latex-display]Example 9.2.L
Factor [latex]25x^{-4}-36[/latex]Answer: Recall that a difference of squares factors in this way: [latex]a^2-b^2=(a-b)(a+b)[/latex], and the first thing we did was identify a and b to see whether we could factor this as a difference of squares. Given [latex]25x^{-4}-36[/latex], we can define [latex]a=5x^{-2},\text{ and }b = 6[/latex] because [latex]{5x^{-2}}^2=25x^{-4},\text{ and }6^2=36[/latex] Therefore the factored form is: [latex](5x^{-2}-6)(5x^{-2}+6)[/latex]
Answer
[latex-display]25x^{-4}-36=(5x^{-2}-6)(5x^{-2}+6)[/latex-display]9.2.4 Fractional Exponents
Again, we will first practice finding a GCF that has a fractional exponent.Example 9.2.M
Factor [latex]x^{\frac{2}{3}}+3x^{\frac{1}{3}}[/latex]Answer: First, look for the term with the lowest value exponent. In this case, it is [latex]3x^{\frac{1}{3}}[/latex]. Recall that when you multiply terms with exponents, you add the exponents. To get [latex]\frac{2}{3}[/latex] you would need to add [latex]\frac{1}{3}[/latex] to [latex]\frac{1}{3}[/latex], so we will need a term whose exponent is [latex]\frac{1}{3}[/latex]. [latex]x^{\frac{1}{3}}\cdot{x^{\frac{1}{3}}}=x^{\frac{2}{3}}[/latex], therefore: [latex-display]x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+3)[/latex-display]
Answer
[latex-display]x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+3)[/latex-display]Example 9.2.N
Factor [latex]25x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49[/latex]Answer: Recall that a perfect square trinomial of the form [latex]a^2+2ab+b^2[/latex] factors as [latex](a+b)^2[/latex] The first step in factoring a perfect square trinomial was to identify a and b. To find a, we ask: [latex](?)^2=25x^{\frac{1}{2}}[/latex], and recall that [latex](x^a)^b=x^{a\cdot{b}}[/latex], therefore we are looking for an exponent for x that when multiplied by 2, will give [latex]\frac{1}{2}[/latex]. You can also think about the fact that the middle term is defined as [latex]2ab[/latex] so a will probably have an exponent of [latex]\frac{1}{4}[/latex], therefore a choice for a may be [latex]5x^{\frac{1}{4}}[/latex] We can check that this is right by squaring a: [latex]{(5x^{\frac{1}{4}})}^{2}=25x^{2\cdot\frac{1}{4}}=25x^{\frac{1}{2}}[/latex] [latex-display]b = 7\text{ and }b^2=49[/latex-display] Now we can check whether [latex]2ab =70x^{\frac{1}{4}}[/latex] [latex-display]2ab=2\cdot{5x^{\frac{1}{4}}}\cdot7=70x^{\frac{1}{4}}[/latex-display] Our terms work out, so we can use the shortcut to factor: [latex-display]25x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49=(5x^{\frac{1}{4}}+7)^2[/latex-display]
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 [latex]\left(5,1\right)[/latex] is a solution to the given system of equations.[latex]\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}[/latex]
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 [latex]x^4+3x^2+2[/latex]Answer: This looks a lot like a trinomial that we know how to factor, [latex]x^2+3x+2=(x+2)(x+1)[/latex], except for the exponents. If we substitute [latex]u=x^2[/latex], and recognize that [latex]u^2=(x^2)^2=x^4[/latex] we may be able to factor this beast! Everywhere there is an [latex]x^2[/latex] we will replace it with a u, then factor. [latex-display]u^2+3u+2=(u+1)(u+2)[/latex-display] 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 [latex]x^2=u[/latex] now that we are done factoring. [latex-display](u+1)(u+2)=(x^2+1)(x^2+2)[/latex-display]
Answer
[latex-display]x^4+3x^2+2=(x^2+1)(x^2+2)[/latex-display]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 [latex]6m^2k-3mk-3k[/latex].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: [latex-display]6m^2k-3mk-3k=3k\left(2m^2-m-1\right)[/latex-display] 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 [latex]2\cdot{-1}=-2[/latex] that sum to [latex]-1[/latex]
Factors of [latex]2\cdot-1=-2[/latex] | Sum of Factors |
---|---|
[latex]2,-1[/latex] | [latex]1[/latex] |
[latex]-2,1[/latex] | [latex]-1[/latex] |
[latex]\left(2m^2-m-1\right)=2m^2-2m+m-1[/latex]
Regroup and find the GCF of each group:[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[/latex]
Now factor [latex](m-1)[/latex] from each term:[latex]2m^2-m-1=(m-1)(2m+1)[/latex]
Don't forget the original GCF that we factored out! Our final factored form is:
[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[/latex]
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.Licenses & Attributions
CC licensed content, Original
- Screenshot: Method to the Madness. Provided by: Lumen Learning License: CC BY: Attribution.
- Image: Shortcut This Way. Provided by: Lumen Learning License: CC BY: Attribution.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Factor Perfect Square Trinomials Using a Formula. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex: Factor a Difference of Squares. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
- Unit 12: Factoring, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex 3: Factor a Sum or Difference of Cubes. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Factor Expressions with Negative Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Factor Expressions with Fractional Exponents. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Factor Expressions Using Substitution. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex: Factoring Polynomials with Common Factors. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.