Using factor theorem, factorize the following:

3x³ + 2x² - 19x + 6

Let, f(x) = 3x³ + 2x² - 19x + 6.

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

f(2) = 3(2)³ + 2(2)² - 19(2) + 6

= 3(8) + 2(4) - 38 + 6

= 24 + 8 - 38 + 6

= 0.

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

Dividing f(x) by (x - 2), we get :

$\begin{array}{l} \phantom{x -.]3)}{3x^2 + 8x - 3} \ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \ \phantom{x - l}\underline{\underset{-}{}3x^3 \underset{+}{-}6x^2} \ \phantom{{x - 2}x^,,,3-}8x^2 - 19x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-}16x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{-3x + 6} \ \phantom{{x - 2}x,'^3o;llk]lmk\space}\underline{\underset{+}{-}3x\underset{-}{+}6} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3)

= (x - 2)(3x² + 9x - x - 3)

= (x - 2)[3x(x + 3) - 1(x + 3)]

= (x - 2)(3x - 1)(x + 3).

Hence, 3x³ + 2x² − 19x + 6 = (x − 2)(3x − 1)(x + 3).