find the roots of x3+6x2+11x+6=0 what is the formula and answer how to solve this ??
Even though a formula to solve a cubic equation exists, nobody in their right mind would use it
http://www.math.vanderbilt.edu/~schectex/courses/cubic/
The first thing I usually do is to try the factor theorem
look for factors of 6 which might work
try x = ±1, ±2, ±3
on my second try, I found x = -1 to be a root,
so x+1 is a factor
Use either long algebraic division or synthetic division to get the other factor of x^2 + 5x + 6
which factors once more to
(x+2)(x+3)
so ...
x^3+6x^2+11x+6=0
(x+1)(x+2)(x+3) = 0
x = -1, -2, -3
We were lucky for this to happen, if no nice rationals were found, we could use more advanced methods such as Newton's Method
To find the roots of the equation x^3 + 6x^2 + 11x + 6 = 0, we can use the Rational Root Theorem or synthetic division. The Rational Root Theorem states that if a rational number p/q is a root of a polynomial equation, then p must be a factor of the constant term and q must be a factor of the leading coefficient.
First, we list all the possible rational roots of the equation. The factors of the constant term 6 are ±1, ±2, ±3, and ±6. The factors of the leading coefficient 1 are ±1. So, the possible rational roots are ±1, ±2, ±3, and ±6.
Using synthetic division with each of the possible rational roots, we can check which ones give a remainder of 0. The root that gives a remainder of 0 is a valid root of the equation. In this case, we find that x = -1 is one of the roots.
Once we have one root, we can use polynomial long division or synthetic division to divide the equation by (x + 1) to get a quadratic equation. By doing this, we obtain the equation x^2 + 5x + 6 = 0.
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, the quadratic equation can be factored as (x + 2)(x + 3) = 0.
Setting each factor equal to zero, we find x + 2 = 0 and x + 3 = 0. Solving these equations, we get x = -2 and x = -3.
Therefore, the roots of the given equation x^3 + 6x^2 + 11x + 6 = 0 are x = -1, x = -2, and x = -3.
To find the roots of a cubic equation of the form ax^3 + bx^2 + cx + d = 0, you can use the Rational Root Theorem and synthetic division. Here's how you can solve the given equation:
Step 1: Identify the coefficients of the equation. In this case, a = 1, b = 6, c = 11, and d = 6.
Step 2: List all possible rational roots that can satisfy the equation using the Rational Root Theorem. The possible roots are ± (factors of d) / (factors of a).
For d = 6 and a = 1, the factors of d are ± 1, ± 2, ± 3, ± 6, and the factor of a is ± 1.
Possible rational roots: ±1, ±2, ±3, ±6
Step 3: Use synthetic division to test each of the possible roots.
Let's start by testing x = 1:
1 | 1 6 11 6
| 1 7 18
-----------------------
1 7 18 24
Since the remainder is not equal to zero, x = 1 is not a root.
Now let's test x = -1:
-1 | 1 6 11 6
| -1 -5 -6
-----------------------
1 5 6 0
Since the remainder is zero, x = -1 is a root.
This means that (x + 1) is a factor of the equation.
Step 4: We can now write the equation as a quadratic equation by dividing it by (x + 1):
(x + 1)(x^2 + 5x + 6) = 0
Expanding the equation:
x^3 + 5x^2 + 6x + x^2 + 5x + 6 = 0
x^3 + 6x^2 + 11x + 6 = 0
Now we have a quadratic equation, x^2 + 5x + 6 = 0, which can be easily solved using the quadratic formula or factoring:
(x + 2)(x + 3) = 0
Setting each factor equal to zero gives us:
x + 2 = 0 or x + 3 = 0
Solving for x:
x = -2 or x = -3
Therefore, the roots of the equation x^3 + 6x^2 + 11x + 6 = 0 are x = -1, x = -2, and x = -3.