32n+1 - 5 x 3n - 2

32n. 3¹ - 3n.5 - 23n (3n.3¹-5) - 2Put 3n = y(3n)².3¹ - (3n).5 - 23y²-5y-23y²+y-6y- 2Y(3y+1)-2(3y+1)(3y+1)(y-2)(3 x 3n+1)(3n-2)→(3n+1 + 1)(3n-2).

Use brackets ! !

Presentation is open to doubt.

3²ⁿ⁺¹ - 5*3ⁿ - 2

= 3*3²ⁿ - 5*3ⁿ - 2

= 3*(3ⁿ)² - 5*3ⁿ - 2

Put y = 3ⁿ, then

3²ⁿ⁺¹ - 5*3ⁿ - 2

= 3*(3ⁿ)² - 5*3ⁿ - 2

= 3y² - 5y - 2

= (3y + 1)(y - 2)

= (3*3ⁿ + 1)(3ⁿ - 2)

= (3ⁿ⁺¹ + 1)(3ⁿ - 2)

Parentheses help. Conventional form for typewritten expressions would make that something like:

3^(2n + 1) + 5*3^n - 2 = 3*3^(2n) - 5*3^n - 2 . . . . . factor a 3 out of the power in the leading term

= 3*(3^n)^2 - 5*3^n - 2 . . . . rewrite 3^(2n) --> 3^(n * 2) = (3^n)^2

= 3u^2 - 5u - 2 . . . above is a polynomial in (3^n). Let u = 3^n and substitute to make that clear

= 3u^2 - 6u + u - 2 . . . . factor by grouping

= (3u)(u - 2) + (u - 2)

= (3u + 1)(u - 2)

= (3*3^n + 1)(3^n - 2) . . . then back-substitute 3^n for u

= [3^(n+1) + 1](3^n - 2) . . .

3^(2n + 1) – 5*3^(n – 2)

= 3^(n - 2) (3^(n + 3) - 5)

3^2n+1 –5*3^n –2 ... did you really mean (2n+1) is the 1st exponent and (n-2) is the 2nd exponent ...

3^(2n+1) – 5*3^(n –2)

= 3^(n –2)[3^(n +1) - 5] <<< answer