what are the roots od x^3+x^2-4x-4?
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i don't understand his example
You may find an explanation you understand in one of these sites.
http://www.google.com/#q=roots+of+x%5E3%2Bx%5E2-4x-4
To find the roots of the cubic polynomial \(x^3+x^2-4x-4\), we can use a technique called factoring or the Rational Root Theorem. Here's how you can find its roots:
Step 1: Setting up the equation
The given equation is \(x^3+x^2-4x-4\). Let's rearrange the terms in descending order:
\(f(x) = x^3 + x^2 - 4x - 4\)
Step 2: Applying the Rational Root Theorem
The Rational Root Theorem states that any rational root of the polynomial must be of the form \(p/q\), where \(p\) is a factor of the constant term (here, -4), and \(q\) is a factor of the leading coefficient (here, 1).
The factors of -4 are \(\pm1, \pm2, \pm4\), and the factors of 1 are \(\pm1\).
Step 3: Testing the possible roots
Now, using the potential rational roots we obtained from the Rational Root Theorem, we can test which ones satisfy the equation
Let's test the roots one by one:
For \(x = -1\):
\(f(-1) = (-1)^3 + (-1)^2 - 4(-1) - 4 = 0\)
Therefore, \(x = -1\) is a root of the equation.
Step 4: Using synthetic division
Divide the original cubic equation by \(x + 1\) using synthetic division to obtain a quadratic equation.
-1 | 1 1 -4 -4
| -1 0 4
|---------------
1 0 -4 0
The result is \(x^2 - 4 = 0\).
Step 5: Factoring the quadratic equation
Now, we have \(x^2 - 4 = 0\).
This quadratic equation can be factored as \((x - 2)(x + 2) = 0\).
Step 6: Solving for x
Setting each factor to zero, we get:
\(x - 2 = 0\) or \(x + 2 = 0\).
Solving these equations, we find:
\(x = 2\) or \(x = -2\).
Hence, the roots of the cubic polynomial \(x^3+x^2-4x-4\) are \(x = -1\), \(x = 2\), and \(x = -2\).