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]2y^{-3}(6-y)=12y^{-3}-2y^{-3+1}=12y^{-3}-2y^{-2}[/latex]

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](x^{-1}+2)(x^{-1}+3)=x^{-1+(-1)}+2x^{-1}+3x^{-1}+6=x^{-2}+5x^{-1}+6[/latex]

The factored form is [latex](x^{-1}+2)(x^{-1}+3)[/latex]

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]

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]x^{\frac{2}{3}}+3x^{\frac{1}{3}}=x^{\frac{1}{3}}(x^{\frac{1}{3}}+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]

[latex]2ab=2\cdot{5x^{\frac{1}{4}}}\cdot7=70x^{\frac{1}{4}}[/latex]

Our terms work out, so we can use the shortcut to factor:

[latex]25y^{\frac{1}{2}}+70x^{\frac{1}{4}}+49=(5x^{\frac{1}{4}}+7)^2[/latex]

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].

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]6m^2k-3mk-3k=3k\left(2m^2-m-1\right)[/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]

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]