Question

factor s^3 + 5s^2 + 9s + 5

Answer 1

It's been so long, I've forgotten how to do these third degree ones

Answer 2

(s+1)(s2+4s+5)

Answer 3

Let p(x)=\[s^3+5s^2+9s+5\]

Answer 4

Then find a zero of the cubic polynomial.

Answer 5

Remember Descartes Rule of Signs and the Rational Root THeorem?

Answer 6

-1 is a zero of this cubic polynomial.
x=-1 ,x+1=0
Therefore,\[s^3+5s^2+9s+5\] is divisible by x+1

Answer 7

Use the rational root theorem to determine possible rational roots.
±5, ±1
Use Descartes Rule of Signs to find the possible possible, negative, and complex roots.
Positive: 0
Negative: 3 or 1. It can't be 3 because there aren't three choices.
x = -3 or -1
Plug it in to synthetic division.

x^2 + 4x + 5 is the depressed equation.
(x + 1)(x^2 + 4x + 5) = 0
Use the quadratic formula to solve for the rest.