Show that (3x + 2) is a factor of (6x³ + 13x² - 4) and hence factorize (6x³ + 13x² - 4).

Let, f(x) = 6x³ + 13x² - 4.

Factor :

⇒ 3x + 2 = 0

⇒ 3x = -2

⇒ x = $-\dfrac{2}{3}$.

Substituting, x = $\Big(\dfrac{-2}{3}\Big)$ in f(x), we get :

$f\Big(-\dfrac{2}{3}\Big) = 6\Big(-\dfrac{2}{3}\Big)^3 + 13\Big(-\dfrac{2}{3}\Big)^2 - 4 \\[1em] = 6\Big(-\dfrac{8}{27}\Big) + 13\Big(\dfrac{4}{9}\Big) - 4 \\[1em] = \Big(-\dfrac{48}{27}\Big) + \Big(\dfrac{52}{9}\Big) - 4 \\[1em] = \dfrac{-48 + 156 - 108}{27} \\[1em] = \dfrac{156 - 156}{27} \\[1em] = 0.$

Since, $f\Big(-\dfrac{2}{3}\Big)$ = 0, thus (3x + 2) is a factor of 6x³ + 13x² - 4.

Now, dividing f(x) by (3x + 2),

$\begin{array}{l} \phantom{x -;]/3)}{2x^2 + 3x - 2} \ 3x + 2\overline{\smash{\big)}6x^3 + 13x^2 - 4} \ \phantom{x - 2l}\underline{\underset{-}{}6x^3 \underset{-}{+}4x^2} \ \phantom{{x - 2}ex^,,,3-}9x^2 \ \phantom{{x e[[-l2}fx^3]\space}\underline{\underset{-}{+}9x^2 \underset{-}{+} 6x} \ \phantom{{x - 2]euo[ki]}x^3ok\space}{-6x - 4} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{+}{-}6x\underset{+}{-} 4} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

6x³ + 13x² - 4 = (3x + 2)(2x² + 3x - 2)

= (3x + 2)(2x² + 4x - x - 2)

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

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

Hence, 6x³ + 13x² - 4 = (3x + 2)(2x - 1)(x + 2).