Essentially, factoring by grouping is nothing more than (maybe) strategically rearranging a quantity, and then dropping parentheses around pairs of added or subtracted terms (terms that have a lot of factors in common), and factoring the GCF out of that quantity. Often, after grouping pairs of terms and factoring out the GCF, one can factor a second time. This is illustrated in the quick example below, and all the examples that follow!
Note that the first two terms \(2x^3\) and \(2x\) have a \(2x\) in common, and that the \(3x^4\) and \(3x^2\) have a \(3x^2\) in common. So, we drop parentheses around these pairs (grouping them) and factor out the GCF from each.
$$\begin{align} 2x^3+2x+3x^4+3x^2 &= (2x^3+2x)+(3x^4+3x^2)\\&=2x(x^2+1)+3x^2(x^2+1)\end{align}$$
Now, notice that we have two terms being added together, namely \(2x(x^2+1)\) and \( 3x^2(x^2+1)\). These both have a factor of \((x^2+1)\) in common. We can factor this out!
$$\begin{align} 2x\cdot (x^2+1)+3x^2\cdot (x^2+1) &=(x^2+1)\cdot[2x+3x^2]\end{align}$$
We could factor this a little further (by factoring out an \(x\) from the \( [2x+3x^2] \), but for now we will refrain from doing so. When factoring by grouping, the hope is that the above sort of thing will happen; where after factoring out GCF from each pair of terms results in two terms where something “large” can be factored out.
Subsequent lessons are about factoring quadratic expressions. Factoring by grouping, as a method, makes this possible and intuitive!
\((2x+3)(6y-10)\)
\((x^2+3)(2-z)\)
\(t(t-8)+1(16-r^2)\)… not much more that can be done. Same goes for any other rearrangement. Thus, not every 4-term polynomial can factor as nicely as the last several examples.
\(2(p^2-2)(q^2+5)\)
\((x+y)(x-2)\)