Finding the Roots of the Function f(x) = x³ + 3x² + x + 3
The Remainder Theorem states that for any polynomial f(x), if you divide it by (x – c), the remainder is f(c). This theorem not only helps in finding the remainder but also indicates if c is a root of the polynomial: if f(c) = 0, then c is a root.
Step 1: Finding Possible Rational Roots
First, we identify the potential rational roots of the polynomial using the Rational Root Theorem. For a polynomial of the form:
f(x) = a_n * x^n + a_{n-1} * x^{n-1} + ... + a_1 * x + a_0
the potential rational roots are the factors of the constant term (in this case, +3) divided by the factors of the leading coefficient (here, +1). Therefore, the possible rational roots are:
- ±1
- ±3
Step 2: Evaluating the Possible Roots
Next, we can evaluate these possible roots by substituting them into f(x):
- For x = 1:
f(1) = 1³ + 3(1)² + 1 + 3 = 1 + 3 + 1 + 3 = 8
So, x = 1 is not a root.
f(-1) = (-1)³ + 3(-1)² + (-1) + 3 = -1 + 3 - 1 + 3 = 4
So, x = -1 is not a root.
f(3) = 3³ + 3(3)² + 3 + 3 = 27 + 27 + 3 + 3 = 60
So, x = 3 is not a root.
f(-3) = (-3)³ + 3(-3)² + (-3) + 3 = -27 + 27 - 3 + 3 = 0
So, x = -3 is indeed a root!
Step 3: Synthetic Division
Now that we have identified -3 as a root, we can perform synthetic division to factor the polynomial:
-3 | 1 3 1 3
| -3 0 -3
-----------------
1 0 1 0
This yields a quotient of x² + 1, meaning we can express the polynomial as:
f(x) = (x + 3)(x² + 1)
The quadratic factor x² + 1 does not have real roots, but we can find its roots in the complex number system.
Step 4: Finding Complex Roots
Setting x² + 1 = 0 gives us:
x² = -1
This results in:
x = i and x = -i
Final Result
Thus, the roots of the function f(x) = x³ + 3x² + x + 3 are:
- x = -3 (real root)
- x = i (complex root)
- x = -i (complex root)
In conclusion, by utilizing the Remainder Theorem and synthetic division, we have successfully determined the roots of the polynomial function.