Using factor theorem, factorize the following:

2x³ + 3x² − 9x − 10

Let, f(x) = 2x³ + 3x² − 9x − 10.

Substituting, x = 2 in f(x), we get :

f(2) = 2(2)³ + 3(2)² − 9(2) − 10

= 16 + 12 − 18 − 10

= 0.

Since, f(2) = 0, (x − 2) is a factor of f(x).

Dividing f(x) by (x − 2), we get :

$\begin{array}{l} \phantom{x - ]k3)}{2x^2 + 7x + 5} \ x - 2\overline{\smash{\big)}2x^3 + 3x^2 - 9x - 10} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \ \phantom{{x - 2}x^,,,3-}7x^2 - 9x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}7x^2 \underset{+}{-}14x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{5x - 10} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}5x\underset{+}{-}10} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 2x³ + 3x² − 9x − 10 = (x − 2)(2x² + 7x + 5)

= (x − 2)(2x² + 5x + 2x + 5)

= (x − 2)[x(2x + 5) + 1(2x + 5)]

= (x − 2)(2x + 5)(x + 1)

Hence, 2x³ + 3x² − 9x − 10 = (x − 2)(2x + 5)(x + 1).