solve this problem please?

Question

please show work

s^3+4s^2-s-4=0


thanks a lot

Answer 1

Factor by grouping
(s^3 + 4s^2) + (-s - 4) = 0
s^2 (s+4) + (-1)(s+4) = 0
(s+4)(s^2 -1) = 0
(s+4)(s+1)(s-1) = 0
s+4 = 0 or s+1 = 0 or s-1 = 0
s = -4, -1 or 1

Answer 2

s^3 + 4s^2 - s - 4 = 0
s^3 + 3s^2 + 3s + 1 + s^2 - 4s - 5 = 0
(s + 1)^3 = - (s^2 - 4s - 5) = -(s + 1)(s - 5)
s^2 + 2s + 1 = - s + 5
s^2 + 3s - 4 = 0
(s + 4)(s - 1) = 0
s^3 + 4s^2 - s - 4 = (s + 1)(s + 4)(s - 1) = 0
s = -4, -1, 1,

-1 + 4 + 1 - 4 = 0
1 + 4 - 1 - 4 = 0
-64 + 64 + 4 - 4 = 0

(s + 4)(s^2 - 1) = 0
s^3 + 4s^2 - s - 4 = 0

Answer 3

given: s^3+4s^2-s-4=0

in general the equation can be written
A*s^3 +B*s^2 + C*s + D = 0...(####)
in our case A=1, D=4

if p/q is a rational root in (####) then
p is divisor in A and q is divisor in D =>

p belongs to {-1,1}
q belongs to {-4,-2,-1,1,2,4}

therefore:
p/q belongs to {-4,-2,-1,1,2,4}

if we try s = p/q = 1 we have
1+4-1-4=0 ergo s = 1 is a root
=======================
if we try s = p/q = -1 we have
-1+4+1-4 = 0 ergo s = -1 is a root
=========================
if we try s= p/q = - 4 we have
-64 + 64 + 4 - 4 = 0 ergo s = -4 is a root
===========================
hence:
s^3 + 4s^2 - s - 4 = ( s-1)(s+1)(s+4) = 0
===========================
Control:
(s-1)(s+1)= s^2-1
(s^2-1)(s+4) = s^3 + 4s^2 - s - 4

Answer 4

Start off by using trial and error.

When s = 1, we see that the equation is satisfied. Therefore (s-1) is a factor of s^3+4s^2-s-4.

Divide s^3+4s^2-s-4 by (s-1) to get (s^2+5s+4)

So,
s^3+4s^2-s-4
= (s-1)(s^2+5s+4)
Factorise further to get
=(s-1)(s+1)(s+4)

Therefore (s-1)(s+1)(s+4)=0 and hence s=1,-1 or -4

Answer 5

s^2(s+4)-(s+4)=0
(s+4)(s^2-1)=0
(s+4)(s-1)(s+1)=0
(s+4)=0
s=-4
(s-1)=0
s=+1
(s+1)=0
s=-1

Answer 6

first graph it on a graphing calculator and find a real root. when u get that synthetically divide by (x- that root) when its down to a quadratic formula, solve by the quadratic formula, then u can find all the roots

hope this helps