Question

n^3-3n^2+2n-1320=0 find n=?......answer the question step by step detailly,I'll mark it as brainliest

Answer 1

Given : n³ - 3n² + 2n - 1320 = 0

⇒ n³ - 12n² + 9n² - 108n + 110n - 1320 = 0

⇒ n³ + 9n² + 110n - 12n² - 108n - 1320 = 0

⇒ n(n² + 9n + 110) - 12(n² + 9n + 110) = 0

⇒ (n - 12)(n² + 9n + 110) = 0

(i)

n - 12 = 0

n = 12

(ii)

n² + 9n + 110 = 0

Here, a = 1, b = 9, c = 110

d = b² - 4ac

  = 81 - 440

  = -359

 

Now,

n = (-b ± √D)/2a

  = (-9 ± √359)/2

  = (-9 ± i√359)/2

Therefore, the values are:

=> n = 12, (-9/2) + (i√359/2), (-9/2) - (i√359)/2

Hope it helps!

Answer 2

Answer:

(-13 ± √359 i) ÷ 2 , 10

Step-by-step explanation:

n^3+3n^2+2n-1320=0

n³ +3n² + 2n -1320=0

n = 10

(  n - 10) ( n² + 13n + 132 ) = 0

n² + 13n + 132  = 0

n = -13 ± √169 - 528 ÷ 2

n = (-13 ± √359 i) ÷ 2 , 10