Question

40
$40n ^{2} + n - 6$

Answer 1

Given :-

$40n {}^{2} + n - 6$
We have to bring out the factors of 40 × 6 such that their algeric sum equals 1

So we have 15 × 16 = 40 × 6

Comming back to equation

$40n {}^{2} + n \: - 6 \\ \\ = 40n {}^{2} + 16n - 15n - 6 \\ \\ = 8n(5n + 2) - 3(5n + 2) \\ \\ = (5n + 2)(8n - 3)$

Hence 40n² can be splitted as ( 5n + 2 ) ( 8n - 3)