Question

5-(3a^2)(6-3a^2+2a) how to solve this ?

Answer 1

$\textbf{Answer :}$

$\textbf{I think that the question be}$

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

So,

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

$ = 5 - m (6 - m) $,
where $ m = 3a^2 - 2a $

$ = 5 - 6m + m^2 $

$ = m^2 - 6m + 5 $

$ = m^2 - m - 5m + 5 $

$ = m (m - 1) - 5 (m - 1) $

$ = (m - 1) (m - 5) $

$ = (3a^2 - 2a - 1) (3a^2 - 2a - 5) $,
since $ m = 3a^2 - 2a $

$ =( (3a^2 - 3a + a - 1) $
$ (3a^2 + 3a - 5a - 5))$

$ = (3a (a - 1) + 1 (a - 1))$
$ (3a (a + 1) - 5 (a + 1)) $

$ = (a - 1) (3a + 1) (a + 1) (3a - 5) $,

$\textbf{which is the required}$
$\textbf{factorization.}$

#$\textbf{MarkAsBrainliest}$

Answer 2

Step-by-step explanation:

it's given above

hope it's helpful for you