(x+1)(x-3)(x-3) = 0
Three real roots
single real at x = -1
double real at x = 3
http://www.wolframalpha.com/input/?i=solve+x%5E3+-5x%5E2+%2B+3x+%2B+9+%3D+0+%3D+0
x^3-5x^2+3x+9=0
Three real roots
single real at x = -1
double real at x = 3
http://www.wolframalpha.com/input/?i=solve+x%5E3+-5x%5E2+%2B+3x+%2B+9+%3D+0+%3D+0
First, we need to guess a possible root of the equation. In this case, let's start with x = 1. We will need to check if it is indeed a root of the equation.
Substitute x = 1 into the equation:
(1)^3 - 5(1)^2 + 3(1) + 9 = 1 - 5 + 3 + 9 = 8
Since the result is not zero, x = 1 is not a root of the equation.
Now, let's try another value. We can use a process called the Rational Root Theorem to find possible rational roots. According to the theorem, the possible rational roots are the divisors of the constant term (9) divided by the divisors of the leading coefficient (1).
The divisors of 9 are ±1, ±3, ±9, and the divisors of 1 are ±1. Combining the possible divisors, we get ±1 and ±9.
We will try these possible rational roots by substituting them one by one into the equation:
1. Substitute x = -1:
(-1)^3 - 5(-1)^2 + 3(-1) + 9 = -1 - 5 - 3 + 9 = 0
Since the result is zero, x = -1 is a root of the equation.
Using synthetic division, divide the equation by (x + 1):
Coefficient of x^3: 1
Coefficient of x^2: -6
Coefficient of x: 9
Constant term: -9
The result of the synthetic division is:
x^2 - 6x + 9
Now, we have a quadratic equation x^2 - 6x + 9 = 0. This can be factored as (x - 3)(x - 3) = 0, where x = 3 is a repeated root. So, the roots of the cubic equation x^3 - 5x^2 + 3x + 9 = 0 are x = -1 (a single root) and x = 3 (a repeated root).