(x%2B3i)(x-3i).jpeg "(x+1)(x+3i)(x-3i)")
Factoring and Solving a Complex Polynomial: (x+1)(x+3i)(x-3i)
This article will explore the factorization and solving of the polynomial (x+1)(x+3i)(x-3i). We will break down the process, discuss the properties of complex numbers, and ultimately arrive at the solutions for this equation.
Understanding Complex Numbers
Complex numbers are numbers that can be expressed in the form a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit, defined as the square root of -1.
Key properties of complex numbers:
- i² = -1: This fundamental property is essential for understanding complex arithmetic.
- Conjugates: Every complex number has a conjugate. The conjugate of a + bi is a - bi. The product of a complex number and its conjugate always results in a real number.
Factoring the Polynomial
The polynomial (x+1)(x+3i)(x-3i) is already factored. This form is useful for finding the roots (solutions) of the equation when set equal to zero.
Finding the Roots (Solutions)
To find the roots, we set the polynomial equal to zero:
(x+1)(x+3i)(x-3i) = 0
For the product of three factors to be zero, at least one of the factors must be zero. This leads to three possible cases:
-
x + 1 = 0
Solving for x, we get x = -1 -
x + 3i = 0
Solving for x, we get x = -3i -
x - 3i = 0 Solving for x, we get x = 3i
The Solutions
Therefore, the solutions (roots) to the equation (x+1)(x+3i)(x-3i) = 0 are:
- x = -1
- x = -3i
- x = 3i
Conclusion
We successfully factored and solved the polynomial (x+1)(x+3i)(x-3i). The solutions are -1, -3i, and 3i. This process demonstrates the use of complex number properties and how to handle equations involving them.