Show that 2x + 7 is a factor of 2x³ + 7x² - 4x - 14. Hence, solve the equation :

2x³ + 7x² - 4x - 14 = 0.

Given,

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = -$\dfrac{7}{2}$

Substituting x = -$\dfrac{7}{2}$ in L.H.S. of 2x³ + 7x² - 4x - 14 = 0, we get :

$\phantom{=} 2 \times \Big(-\dfrac{7}{2}\Big)^3 + 7 \times \Big(-\dfrac{7}{2}\Big)^2 - 4 \times -\dfrac{7}{2} - 14 \\[1em] = 2 \times -\dfrac{343}{8} + 7 \times \dfrac{49}{4} + 14 - 14 \\[1em] = -\dfrac{343}{4} + \dfrac{343}{4} + 14 - 14 \\[1em] = 0.$

Since, L.H.S. = R.H.S. = 0.

∴ 2x + 7 is a factor of 2x³ + 7x² - 4x - 14 = 0.

Dividing 2x³ + 7x² - 4x - 14 = 0 by 2x + 7:

$\begin{array}{l} \phantom{2x + 7)}{x^2 - 2} \ 2x + 7\overline{\smash{\big)}2x^3 + 7x^2 - 4x - 14} \ \phantom{2x + 7}\underline{\underset{-}{}2x^3 \underset{-}{+}7x^2} \ \phantom{{2x + 7}2x^3} \times \phantom{+25^2} - 4x - 14 \ \phantom{{2x + 7}x^3+2+25^2}\underline{\underset{+}{-}4x \underset{+}{-} 14} \ \phantom{{2x + 7}x^3+2-5x^23-3}\times \end{array}$

∴ 2x³ + 7x² - 4x - 14 = (2x + 7)(x² - 2)

= (2x + 7)(x + $\sqrt{2})(x - \sqrt{2}$)

Solving,

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = -$\dfrac{7}{2}$

Also,

⇒ x + $\sqrt{2}$ = 0

⇒ x = -$\sqrt{2}$ = -1.41

Also,

⇒ x - $\sqrt{2}$ = 0

⇒ x = $\sqrt{2}$ = 1.41

Hence, x = -$\dfrac{7}{2}$, 1.41, -1.41