You can't. 17 + 2n = 21 + 2n gives: 2n = 4 + 2n which gives 0=4 which is not possible so the sum is not solveable.
Exactly that. 2n+2.
(2n + 7)(2n + 7)
n is 2. To solve, do the division: . . . . . . . . .x2 +. 3x + (6-2n) . . . -------------------------- x-2 | x3 + x2 - 2nx + n2 . . . . .x3 -2x2 . . . . .-------- . . . . . . . .3x2 - 2nx . . . . . . . .3x2 - . 6x . . . . . . . .---------- . . . . . . . . . (6-2n)x + n2 . . . . . . . . . (6-2n)x - 2(6-2n) . . . . . . . . . ---------------------- . . . . . . . . . . . . . . . .n2 + 2(6-2n) But this remainder is known to be 8, so: n2 + 2(6-2n) = 8 ⇒ n2 - 4n + 4 = 0 ⇒ (n - 2)2 = 0 ⇒ n = 2
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
That factors to (2n + 1)(n + 2)
n(2n - 1)(2n + 7)
Oh, dude, adding n squared plus n squared is like adding apples to apples, you know? It's just like, you take two n squared terms and you add them together to get 2n squared. It's not rocket science, man. Just double up those n squares and you're good to go.
6n2 + 16n can be factorised to give 2n(3n + 8) so the highest factor is 2n.
If the given equation were factorized it would be: 2n2 + 4n = 2n(n+1) Another Answer:- The given expression when factored is 2n(n+2)
(n2 + 2n - 1) (n2 + 2n - 1) = n4 + 2n3 - n2 + 2n3 + 4n2 - 2n - n2 - 2n + 1 = n4 + 4n3 + 2n2 - 4n + 1 try with n = 5: (5 squared + 10 - 1) squared = 34 squared = 1156 with formula (5^4) + (4 *(5^3)) + (2 * (5^2)) - (4 * 5) + 1 = 625 + 500 + 50 - 20 + 1 = 1156
2n + 4m - 2n + m = 5m
2n plus 4 = 2n + 4
4n
7+2n
It is: 2n^2 -5n -42 equals (2n+7)(n-6) when factored
It is 2n^3 + 7n^2 - 13n + 3