Read: More Factoring Methods
Learning Objectives
- Factor expressions with negative exponents
- Factor expressions with fractional exponents
- Factor by substitution
- Factor completely
- 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
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
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 [latex]1[/latex], 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
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]Fractional Exponents
Again, we will first practice finding a GCF that has a fractional exponent.Example
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
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 [latex]2[/latex], 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]
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
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]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
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 [latex]3k[/latex]. Factor [latex]3k[/latex] 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 [latex]4th[/latex] degree polynomial. The last topic we covered was what it means to factor completely.Licenses & Attributions
CC licensed content, Original
- 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.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Factor Expressions with Negative Exponents. 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) . 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.
