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Study Guides > Intermediate Algebra

Read: More Factoring Methods

Learning Objectives

  • Factor expressions with negative exponents
  • Factor expressions with fractional exponents
  • Factor by substitution
  • Factor completely
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: x1x1=x1+(1)=x2x^{-1}\cdot{x^{-1}}=x^{-1+(-1)}=x^{-2}
  • When you add fractions, you need a common denominator: 12+13=3312+2213=36+26=56\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

Factor 12y32y212y^{-3}-2y^{-2}

Answer: If the exponents in this expression were positive we could determine that the GCF is 2y22y^2, but since we have negative exponents, we will need to use 2y22y^{-2}. Therefore 12y32y2=2y2(6y11)12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1) We can check that we are correct by multiplying: 2y2(6y11)=12y2+(1)2y2=12y32y22y^{-2}(6y^{-1}-1)=12y^{-2+(-1)}-2y^{-2}=12y^{-3}-2y^{-2}

Answer

12y32y2=2y2(6y11)12y^{-3}-2y^{-2}=2y^{-2}(6y^{-1}-1)

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

Example

Factor x2+5x1+6x^{-2}+5x^{-1}+6.

Answer: If the exponents on this trinomial were positive, we could factor this as (x+2)(x+3)(x+2)(x+3).  Note that the exponent on the x's in the factored form is 11, in other words (x+2)=(x1+2)(x+2)=(x^{1}+2). Also note that 1+(1)=2-1+(-1) = -2, therefore if we factor this trinomial as (x1+2)(x1+3)(x^{-1}+2)(x^{-1}+3), we will get the correct result if we check by multiplying. (x1+2)(x1+3)=x1+(1)+2x1+3x1+6=x2+5x1+6(x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6

Answer

x2+5x1+6=(x1+2)(x1+3)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

Factor 25x43625x^{-4}-36

Answer: Recall that a difference of squares factors in this way: a2b2=(ab)(a+b)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 25x43625x^{-4}-36, we can define a=5x2, and b=6a=5x^{-2},\text{ and }b = 6 because (5x2)2=25x4, and 62=36({5x^{-2}})^2=25x^{-4},\text{ and }6^2=36 Therefore the factored form is: (5x26)(5x2+6)(5x^{-2}-6)(5x^{-2}+6)

Answer

25x436=(5x26)(5x2+6)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

 Fractional Exponents

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

Example

Factor x23+3x13x^{\frac{2}{3}}+3x^{\frac{1}{3}}

Answer: First, look for the term with the lowest value exponent.  In this case, it is 3x133x^{\frac{1}{3}}. Recall that when you multiply terms with exponents, you add the exponents. To get 23\frac{2}{3} you would need to add 13\frac{1}{3} to 13\frac{1}{3}, so we will need a term whose exponent is 13\frac{1}{3}. x13x13=x23x^{\frac{1}{3}}\cdot{x^{\frac{1}{3}}}=x^{\frac{2}{3}}, therefore: x23+3x13=x13(x13+3)x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+3)

Answer

x23+3x13=x13(x13+3)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

Factor 25x12+70x14+4925x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49

Answer: Recall that a perfect square trinomial of the form a2+2ab+b2a^2+2ab+b^2 factors as (a+b)2(a+b)^2 The first step in factoring a perfect square trinomial  was to identify a and b. To find a, we ask:  (?)2=25x12(?)^2=25x^{\frac{1}{2}}, and recall that (xa)b=xab(x^a)^b=x^{a\cdot{b}}, therefore we are looking for an exponent for x that when multiplied by 22, will give 12\frac{1}{2}.  You can also think about the fact that the middle term is defined as 2ab2ab so a will probably have an exponent of 14\frac{1}{4}, therefore a choice for a may be 5x145x^{\frac{1}{4}} We can check that this is right by squaring a: (5x14)2=25x214=25x12{(5x^{\frac{1}{4}})}^{2}=25x^{2\cdot\frac{1}{4}}=25x^{\frac{1}{2}} b=7 and b2=49b = 7\text{ and }b^2=49 Now we can check whether 2ab=70x142ab =70x^{\frac{1}{4}} 2ab=25x147=70x142ab=2\cdot{5x^{\frac{1}{4}}}\cdot7=70x^{\frac{1}{4}} Our terms work out, so we can use the shortcut to factor: 25x12+70x14+49=(5x14+7)225x^{\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

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 (5,1)\left(5,1\right) is a solution to the given system of equations.
x+3y=82x9=y\begin{array}{l}x+3y=8\hfill \\ 2x - 9=y\hfill \end{array}
We can substitute the ordered pair (5,1)\left(5,1\right) into both equations.

(5)+3(1)=8 8=8True2(5)9=(1) 1=1True\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

Factor x4+3x2+2x^4+3x^2+2

Answer: This looks a lot like a trinomial that we know how to factor x2+3x+2=(x+2)(x+1)x^2+3x+2=(x+2)(x+1) except for the exponents. If we substitute u=x2u=x^2, and recognize that u2=(x2)2=x4u^2=(x^2)^2=x^4 we may be able to factor this beast! Everywhere there is an x2x^2 we will replace it with a u, then factor. u2+3u+2=(u+1)(u+2)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 x2=ux^2=u now that we are done factoring. (u+1)(u+2)=(x2+1)(x2+2)(u+1)(u+2)=(x^2+1)(x^2+2)

Answer

x4+3x2+2=(x2+1)(x2+2)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

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 6m2k3mk3k6m^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 3k3k. Factor 3k3k from the trinomial: 6m2k3mk3k=3k(2m2m1)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 21=22\cdot{-1}=-2 that sum to 1-1

Factors of 21=22\cdot-1=-2 Sum of Factors
2,12,-1 11
2,1-2,1 1-1
Our factors are 2,1-2,1, so we can factor by grouping: Rewrite the middle term with the factors we found with the table:

(2m2m1)=2m22m+m1\left(2m^2-m-1\right)=2m^2-2m+m-1

Regroup and find the GCF of each group:

(2m22m)+(m1)=2m(m1)+1(m1)(2m^2-2m)+(m-1)=2m(m-1)+1(m-1)

Now factor (m1)(m-1) from each term:

2m2m1=(m1)(2m+1)2m^2-m-1=(m-1)(2m+1)

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

6m2k3mk3k=3k(m1)(2m+1)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 4th4th degree polynomial.  The last topic we covered was what it means to factor completely.

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