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أدلة الدراسة > ALGEBRA / TRIG I

More Factoring Methods

Learning Outcomes

  • 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].

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^{-3}[/latex]. Therefore, [latex]12y^{-3}-2y^{-2}=2y^{-3}(6-y)[/latex] We can check that we are correct by multiplying: [latex-display]2y^{-3}(6-y)=12y^{-3}-2y^{-3+1}=12y^{-3}-2y^{-2}[/latex-display]

Try It

[ohm_question]93663[/ohm_question]
Now let us factor a trinomial that has negative exponents.

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] 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].

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]

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

 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]

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

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 [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-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]

Try It

[ohm_question]93668[/ohm_question]
In our next video, you will see more examples of how to factor expressions with fractional exponents. https://youtu.be/R6BzjR2O4z8

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 equation:
[latex]x+3y=8[/latex]
To determine whether [latex]x=5[/latex], and [latex]y=1[/latex] is a solution to the equation, we can substitute the values [latex]x=5[/latex] and [latex]y=1[/latex] into the equation.

[latex](5)+3(1)=8[/latex]

[latex]8=8[/latex]     True

We replaced the variables with numbers 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]. 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-display]u^2+3u+2=(u+1)(u+2)[/latex-display] 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-display](u+1)(u+2)=(x^2+1)(x^2+2)[/latex-display] We conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[/latex].

Try It

[ohm_question]1366[/ohm_question]
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

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.

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 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. 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 [latex]4th[/latex] degree polynomial. The last topic we covered was what it means to factor completely.

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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.