Solve the following equation by factorization:

5(3x + 1)² + 6(3x + 1) - 8 = 0

Let us consider y = 3x + 1.

Substituting y = 3x + 1 in equation 5(3x + 1)² + 6(3x + 1) - 8 = 0, we get :

⇒ 5y² + 6y - 8 = 0

⇒ 5y² + 10y - 4y - 8 = 0

⇒ 5y(y + 2) - 4(y + 2) = 0

⇒ (5y - 4)(y + 2) = 0

⇒ (5y - 4) = 0 or (y + 2) = 0      [Using Zero-product rule]

⇒ 5y = 4 or y = -2

⇒ y = $\dfrac{4}{5}$ or y = -2.

Now we have,

Case 1 : y = $\dfrac{4}{5}$

$\Rightarrow y = \dfrac{4}{5} \\[1em] \Rightarrow 3x + 1 = \dfrac{4}{5} \\[1em] \Rightarrow 5(3x + 1) = 4 \\[1em] \Rightarrow 15x + 5 = 4 \\[1em] \Rightarrow 15x = 4 - 5 \\[1em] \Rightarrow 15x = -1 \\[1em] \Rightarrow x = \dfrac{-1}{15}.$

Case 2 : y = -2

⇒ y = -2

⇒ 3x + 1 = -2

⇒ 3x = -2 - 1

⇒ 3x = -3

⇒ x = $\dfrac{-3}{3}$

⇒ x = -1.

Hence, x = $\Big{-1, \dfrac{-1}{15}\Big}$.