Factorise :

$(a^2 + 3a - 5) (a^2 + 3a + 2) + 6$

Given:

$(a^2 + 3a - 5) (a^2 + 3a + 2) + 6$

Let $a^2 + 3a$ be t.

$= (t - 5)(t + 2) + 6\\[1em] = t^2 - 5t + 2t - 10 + 6\\[1em] = t^2 - 3t - 4\\[1em] = t^2 - 4t + t - 4\\[1em] = t(t - 4) + 1(t - 4)\\[1em] = (t - 4)(t + 1)\\[1em]$

Substituting the value of t, we get:

$= [a^2 + 3a - 4](a^2 + 3a + 1)\\[1em] = [a^2 + 4a - a - 4](a^2 + 3a + 1)\\[1em] = [a(a + 4) - 1(a + 4)](a^2 + 3a + 1)\\[1em] = (a + 4)(a - 1)(a^2 + 3a + 1)$

Hence, $(a^2 + 3a - 5) (a^2 + 3a + 2) + 6 = (a + 4)(a - 1)(a^2 + 3a + 1)$.