<PRE><B>Question 241098</B>
There are a variety of factoring techniques:&lt;ul&gt;&lt;li&gt;Greatest Common Factor (GCF)&lt;/li&gt;&lt;li&gt;Patterns&lt;ul&gt;&lt;li&gt;Difference of Squares: {{{a^2 - b^2 = (a+b)(a-b)}}}&lt;/li&gt;&lt;li&gt;Difference of Cubes: {{{a^3 - b^3 = (a-b)(a^2 +ab +b^2)}}}&lt;/li&gt;&lt;li&gt;Sum of Cubes: {{{a^3 + b^3 = (a+b)(a^2 -ab +b^2)}}}&lt;/li&gt;&lt;li&gt;Perfect Trinomial Squares:&lt;ul&gt;&lt;li&gt;{{{a^2 + 2ab + b^2 =  (a+b)(a+b) = (a+b)^2}}}&lt;/li&gt;&lt;li&gt;{{{a^2 - 2ab + b^2 =  (a-b)(a-b) = (a-b)^2}}}&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;li&gt;Trinomial factoring {{{ax^2 + bx + c}}}&lt;/li&gt;&lt;li&gt;Factoring by grouping&lt;/li&gt;&lt;li&gt;Factoring by trial and error of the possible rational roots&lt;/li&gt;&lt;/ul&gt;Always start with GCF factoring. After that, be ready to use any of the other techniques in any order and possibly using some of them repeatedly.&lt;br&gt;
{{{8x^4-27x+16x^3-54}}}
The GCF of this expression is 1 and this is the only GCF we do not often factor out. Next I would look at factoring by grouping because&lt;ul&gt;&lt;li&gt;There are 4 terms which is too many for any of the patterns or for trinomial factoring.&lt;/li&gt;&lt;li&gt;The odd order of the terms is possibly a clue that factoring by grouping should be done.&lt;/li&gt;&lt;li&gt;Factoring by trial and error of the possible rational roots looks like it would be tedious because of the large number of possible rational roots.&lt;/li&gt;&lt;/ul&gt;
When factoring by grouping you factor GCF&#39;s from pairs (or sets) of terms. (Note: This is one of the rare times where you &lt;i&gt;do&lt;/i&gt; factor out a GCF that is one!)
{{{green(x)\red((8x^3 - 27)) + green(2)red((8x^3-27))}}}&lt;br&gt;
Now, if we are lucky, the &quot;other&quot; factors (i.e. the factors that were not the GCF of the pair, the ones in red) in each group are the same. And as you can see, we are lucky. The {{{(8x^3-27)}}}&#39;s are the same. (If these &quot;other factors&quot; did not match, then we would start over, rearrange the terms and try again. If you run out of rearrangements and never get matching &quot;other factors&quot; then the expression will not factor with this technique.)&lt;br&gt;
Now we factor out the matching &quot;other factors&quot;. This is perhaps the most difficult step to learn when factoring by grouping so I have color-coded the before and after expressions so you can see where things were and where they end up:
{{{red((8x^3 - 27))green((x + 2))}}}&lt;br&gt;
When you reduce fractions you always reduce them until you can reduce them any further. Factoring is like this, too. Always keep factoring until you can&#39;t factor any more.&lt;br&gt;
Since {{{8x^3 = (2x)^3}}} and {{{27 = 3^3}}}, the first factor fits the &quot;Difference of Cubes&quot; pattern which we can use to factor it. (The last factor, (x+2), will not factor further.)
{{{((2x) - (3))((2x)^2 + (2x)(3) + (3)^2)(x+2)}}}
which simplifies to:
{{{(2x - 3)(4x^2 + 6x + 9)(x+2)}}}
The first factor will not factor further. The second factor looks like it might be factorable. But, as long as you have already factored out the GCF of the entire expression (which you should have done before anything else), then the second factor from the Difference (or Sum) of Cubes is inherently unfactorable. If you know this then you will know not to waste time trying.&lt;br&gt;
Since none of the factors will factor any further we are finished!</PRE>