A Missing Middle

$x^{2}-9$ has no middle term. In the last section you expanded $\left(x+3\right)\left(x-3\right)$. The cross products $3x$ and $-3x$ cancelled, and the result was $x^{2}-9$. Factoring goes the other way. You start from $x^{2}-9$ and write the two factors back.

A difference of squares is $x^{2}$ minus a square number. It factors into a sum and a difference of the same two terms: $x^{2}-9 = \left(x+3\right)\left(x-3\right)$. The 3 is the square root of 9. There is no middle term because the cross products cancelled.

You can also find the same answer with factor pairs. The numbers 3 and -3 have product -9 and sum 0. So $x^{2}-9 = \left(x+3\right)\left(x-3\right)$. This is the sum and product method with a middle coefficient of 0.