Notice that [latex]25{x}^{2}[/latex] and [latex]4[/latex] are perfect squares because [latex]25{x}^{2}={\left(5x\right)}^{2}[/latex] and [latex]4={2}^{2}[/latex]. Then check to see if the middle term is twice the product of [latex]5x[/latex] and [latex]2[/latex]. The middle term is, indeed, twice the product: [latex]2\left(5x\right)\left(2\right)=20x[/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(5x+2\right)}^{2}[/latex].

Notice that [latex]9{x}^{2}[/latex] and [latex]25[/latex] are perfect squares because [latex]9{x}^{2}={\left(3x\right)}^{2}[/latex] and [latex]25={5}^{2}[/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\left(3x+5\right)\left(3x - 5\right)[/latex].

Notice that [latex]{x}^{3}[/latex] and [latex]512[/latex] are cubes because [latex]{8}^{3}=512[/latex]. Rewrite the sum of cubes as [latex]\left(x+8\right)\left({x}^{2}-8x+64\right)[/latex].

Analysis of the Solution

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

Notice that [latex]8{x}^{3}[/latex] and [latex]125[/latex] are cubes because [latex]8{x}^{3}={\left(2x\right)}^{3}[/latex] and [latex]125={5}^{3}[/latex]. Write the difference of cubes as [latex]\left(2x - 5\right)\left(4{x}^{2}+10x+25\right)[/latex].

Analysis of the Solution

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

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].

This looks a lot like a difference of squares that we know how to factor: [latex]x^2-81=(x+9)(x-9)[/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 will be able to factor.

Everywhere there is a [latex]x^2[/latex] we will replace it with a [latex]u[/latex] then factor.

[latex]u^2-81=(u+9)(u-9)[/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].

[latex](u+9)(u-9)=(x^2+9)(x^2-9)[/latex]

However, this is not fully factored. We notice that [latex]x^2-9[/latex] can be factored into [latex](x+3)(x-3)[/latex]. You always want to factor as much as possible.

Therefore, we conclude that [latex]x^4-81=(x^2+9)(x+3)(x-3)[/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](2m^2-m-1)=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]

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]2[/latex].

Factor [latex]2[/latex] from the binomial:

[latex]2(x^4-72x^2+1296)[/latex]

We are left with a trinomial that can be factored using perfect squares. Notice that [latex]x^4=(x^2)^2[/latex] and [latex](36)^2=1296[/latex]. Also notice that the middle term, [latex]-72x^2[/latex], of the trinomial is equal to [latex]-2[/latex] times the product of [latex]x^2[/latex] and [latex]36[/latex]. Therefore, [latex]-2 \cdot x^2 \cdot 36=-72x^2[/latex]. We can conclude that the trinomial [latex]x^4-72x^2+1296[/latex] is a perfect square trinomial.

Since the middle term is negative, use the formula [latex]A^2-2AB+B^2=(A-B)^2[/latex] to factor the perfect square trinomial.
[latex]x^4-72x^2+1296=(x^2)^2-2 \cdot x^2 \cdot 36+(36)^2=(x^2-36)^2[/latex].

Notice that [latex]x^2-36[/latex] is a difference of two squares. We can factor this into [latex](x+6)(x-6)[/latex]. Since this is squared, then [latex](x^2-36)^2=[(x+6)(x-6)]^2=(x+6)^2(x-6)^2[/latex].

Putting this all together, [latex]2x^4-144x^2+2592=2(x+6)^2(x-6)^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]2x^{14}[/latex].

Factor [latex]2x^{14}[/latex] from the binomial:

[latex]2x^{14}(27x^3+1)[/latex]

We are left with a binomial that can be factored using the sum of cubes. Notice that [latex]27x^3[/latex] and 1 are cubes because [latex]27x^3=(3x)^3[/latex] and [latex]1=1^3[/latex]. Therefore, [latex]27x^3+1=(3x+1)(9x^2-3x+1)[/latex]

Our final answer must include the original GCF we already factored out: [latex]2x^{14}(3x+1)(9x^2-3x+1)[/latex]