Solution 2.1:3e

Both terms contain \displaystyle x, which can therefore be taken out as a factor (as can 2).

\displaystyle 18x-2x^3=2x\cdot 9-2x \cdot x^2=2x(9-x).

The remaining second-degree factor \displaystyle 9-x^2 can then be factorized using the conjugate rule

\displaystyle 2x(9-x^2)=2x(3^2-x^2)=2x(3+x)(3-x),

which can also be written as \displaystyle -2x(x+3)(x-3).