How to find the roots of s^3 + 6s^2 + 18s + 20 =0 ???

[From: ] [author: ] [Date: 11-10-31] [Hit: ]

Rearrange the terms of the LHS to get a factor s + 2 as follows : LHS = s³ + 6s² +18s + 20 =s³ + 2s² + 4s² + 8s + 10s + 20 = (s³ + 2s²) + (4s² + 8s) + (10s + 20) = s²(s + 2) + 4s(s + 2) + 10(s + 2) = (s + 2)(s² + 4s + 10) Do you understand now?-s^3 + 6s^2 + 18s + 20 =0s(s^2 + 6s + 18 + 20) = 0s(s^2 + 6s) = - 381. s = -382. s^2 + 6ss(s + 6) = 0s = 03.......

Substituting the variables back, yields :
t = cuberoot(u) = 1.00000000.
z = 0.00000000.
y = 0.00000000.
x = -2.00000000.
The other roots can be found by dividing and solving the remaining quadratic equation.
The other roots are complex : -2.00000000 +- 2.44948974 i.

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s³ + 6s² +18s + 20 = 0
By trial find out what value of s make LHS = 0
Start with s = 1, - 1, 2, - 2 etc.
We find that when s = - 2 , LHS = 0
Hence s + 2 = 0 is afactor of the LHS expr.
Rearrange the terms of the LHS to get a factor s + 2 as follows :
LHS = s³ + 6s² +18s + 20
= s³ + 2s² + 4s² + 8s + 10s + 20
= (s³ + 2s²) + (4s² + 8s) + (10s + 20)
= s²(s + 2) + 4s(s + 2) + 10(s + 2)
= (s + 2)(s² + 4s + 10)
Do you understand now?

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s^3 + 6s^2 + 18s + 20 =0

s(s^2 + 6s + 18 + 20) = 0

s(s^2 + 6s) = - 38

1. s = -38

2. s^2 + 6s

s(s + 6) = 0

s = 0

3. s + 6 =0

s = - 6

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