Factor by Grouping

Four terms with no factor common to all of them can still factor. Take ${x}^{3}+2{x}^{2}+5x+10$. Split it into two pairs and factor each pair on its own: ${x}^{3}+2{x}^{2}$ is ${x}^{2}\left(x+2\right)$, and $5x+10$ is $5\left(x+2\right)$.

Both pairs left the same bracket. Now $x+2$ is a common factor of the two parts, just as y is in 3y + 5y, so take it out:

$${x}^{2}\left(x+2\right)+5\left(x+2\right) = \left(x+2\right)\left({x}^{2}+5\right)$$

Grouping works only when the two brackets match; here they do because 1 and 2 are in the same ratio as 5 and 10. And ${x}^{2}+5$ has no real factors: for real x it is never zero.