Factor polynomials with negative or fractional exponents
Factor by substitution
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.
First, practice finding a GCF that is a negative exponent.
Example
Factor [latex]12y^{-3}-2y^{-2}[/latex].
Show Solution
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^{-3}[/latex].
Now let us factor a trinomial that has negative exponents.
Example
Factor [latex]x^{-2}+5x^{-1}+6[/latex].
Show Solution
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.
The factored form is [latex](x^{-1}+2)(x^{-1}+3)[/latex]
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
Factor [latex]25x^{-4}-36[/latex].
Show Solution
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]
In the following video, you will see more examples that are similar to the previous three written examples.
Fractional Exponents
Again, we will first practice finding a GCF that has a fractional exponent.
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].
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=25y^{\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 [latex]a[/latex] will probably have an exponent of [latex]\frac{1}{4}[/latex], therefore a choice for [latex]a[/latex] may be [latex]5x^{\frac{1}{4}}[/latex].
We can check that this is right by squaring [latex]a[/latex]: [latex]{(5x^{\frac{1}{4}})}^{2}=25x^{2\cdot\frac{1}{4}}=25x^{\frac{1}{2}}[/latex]
[latex]b = 7\text{ and }b^2=49[/latex]
Now we can check whether [latex]2ab =70x^{\frac{1}{4}}[/latex]
In our next video, you will see more examples of how to factor expressions with fractional exponents.
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 is not 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, consider the following system of equations:
To determine whether the ordered pair [latex]\left(5,1\right)[/latex] is a solution to the given system of equations, we can substitute the ordered pair [latex]\left(5,1\right)[/latex] into both equations.
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].
Show Solution
This looks a lot like a trinomial that we know how to factor: [latex]x^2+3x+2=(x+2)(x+1)[/latex]. The only thing different is 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 a [latex]x^2[/latex] we will replace it with a [latex]u[/latex] then factor.
[latex]u^2+3u+2=(u+1)(u+2)[/latex]
We are not quite done yet. We want to factor the original polynomial which had [latex]x[/latex] as its variable, so we need to replace [latex]x^2=u[/latex] now that we are done factoring.
[latex](u+1)(u+2)=(x^2+1)(x^2+2)[/latex]
We conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[/latex].
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.
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 [latex]6m^2k-3mk-3k[/latex] completely.
Show Solution
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].
We are left with a trinomial that can be factored using your choice of factoring methods. 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]
Our factors are [latex]-2,1[/latex] which will allow us to factor by grouping:
Rewrite the middle term with the factors we found:
[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]
Do not forget the original GCF that we factored out! Our final factored form is:
[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[/latex]
In our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this module.
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.