Example

Factor $12y^{-3}-2y^{-2}$.

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

Example

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

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

Example

Factor $25x^{-4}-36$.

Answer: Recall that a difference of squares factors in this way: $a^2-b^2=(a-b)(a+b)$, and the first thing we did was identify a and b to see whether we could factor this as a difference of squares. Given $25x^{-4}-36$, we can define $a=5x^{-2} \text{ and }b = 6$ because $({5x^{-2}})^2=25x^{-4} \text{ and }6^2=36$ Therefore, the factored form is: $(5x^{-2}-6)(5x^{-2}+6)$

Example

Factor $x^{\frac{2}{3}}+3x^{\frac{1}{3}}$.

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

Example

Factor $25x^{\frac{1}{2}}+70x^{\frac{1}{4}}+49$.

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

Example

Factor $x^4+3x^2+2$.

Answer: This looks a lot like a trinomial that we know how to factor: $x^2+3x+2=(x+2)(x+1)$. The only thing different is the exponents. If we substitute $u=x^2$ and recognize that $u^2=(x^2)^2=x^4$, we may be able to factor this beast! Everywhere there is a $x^2$ we will replace it with a $u$ then factor. $$u^2+3u+2=(u+1)(u+2)$$ We are not quite done yet. We want to factor the original polynomial which had $x$ as its variable, so we need to replace $x^2=u$ now that we are done factoring. $$(u+1)(u+2)=(x^2+1)(x^2+2)$$ We conclude that $x^4+3x^2+2=(x^2+1)(x^2+2)$.

Example

Factor $6m^2k-3mk-3k$ 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 $3k$. Factor $3k$ from the trinomial: $$6m^2k-3mk-3k=3k\left(2m^2-m-1\right)$$ We are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of $2\cdot{-1}=-2$ that sum to $-1$

Factors of $2\cdot-1=-2$	Sum of Factors
$2,-1$	$1$
$-2,1$	$-1$

Our factors are $-2,1$ which will allow us to factor by grouping: Rewrite the middle term with the factors we found:

$\left(2m^2-m-1\right)=2m^2-2m+m-1$

Regroup and find the GCF of each group:

$(2m^2-2m)+(m-1)=2m(m-1)+1(m-1)$

Now factor $(m-1)$ from each term:

$2m^2-m-1=(m-1)(2m+1)$

Do not forget the original GCF that we factored out! Our final factored form is:

$6m^2k-3mk-3k=3k (m-1)(2m+1)$