Question

Add the sum of (6m ^ 2 - n ^ 2 + 1) and (4m ^ 2 - n ^ 2) to the product of (6m + n) and (2m - 3n)
*answer in formula pls*​

Answer 1

Answer:

$= {10m}^{2} - {5n}^{2} - 20mn + 12m + 1$

Step-by-step explanation:

$sum \: of \: ({6m}^{2} - {n}^{2} + 1) \: and \: ({4m}^{2} - {n}^{2} )$

$= {6m}^{2} - {n}^{2} + 1 + {4m}^{2} - {n}^{2} \\ = {10m}^{2} - {2n}^{2} + 1 - - - - > (1)$

$product \: of \: (6m + n) \: and \: (2m - 3n)$

$= 6m(2m - 3n) + n(2m - 3n) \\ = 12m - 18mn + 2mn - {3n}^{2} \\ = 12m - 20mn - {3n}^{2} - - - - > (2)$

$adding \: (1) \: and \: (2) \\ = {10m}^{2} - {2n}^{2} + 1 + 12m - 20mn - {3n}^{2} \\ = {10m}^{2} - {5n}^{2} - 20mn + 12m + 1$