Let’s resolve into factors

x²-bx-(a + 3b)(a + 2b)

x²-bx-(a + 3b)(a + 2b) … (1)

Given, x² + (p + q)x + pq=(x + p)(x + q) … (2)

Comparing 1 and 2

p + q =-b … (3)

pq =-(a + 3b)(a + 2b) … (4)

to make the factors of 3 from 4

-b= (a + 2b) – (a + 3b)

∴ (1) can be written as

x² + (a + 2b)x – (a + 3b)x-(a + 3b)(a + 2b)

=[{x + (a + 2b)}{x-(a + 3b)}]

∴ the resolved factors are [{x + a + 2b)}{x- a - 3b)}]