please explain the process of finding the complex zeros of f(x)=x^4+5x^2+4.
so we are solving x^4+5x^2+4
it factors
(x^2+1)(x^2+4) = 0
x^2 = -1 or x^2 = -4
x = +/- i or x = +/- 2i where i = √-1
Thank you
To find the complex zeros of f(x) = x^4 + 5x^2 + 4, we can use the fact that a complex number is a zero of a polynomial if and only if its conjugate is also a zero.
Step 1: Rewrite the polynomial in terms of a quadratic equation.
Let's substitute x^2 = t. This will help us simplify the equation and make it easier to solve.
So, our equation becomes: t^2 + 5t + 4 = 0.
Step 2: Solve the quadratic equation.
To solve the quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = 1, b = 5, and c = 4. Plugging these values into the quadratic formula, we get:
t = (-5 ± √(5^2 - 4 * 1 * 4)) / (2 * 1)
t = (-5 ± √(25 - 16)) / 2
t = (-5 ± √9) / 2
t = (-5 ± 3) / 2
Thus, we have two possible values for t:
t1 = (-5 + 3) / 2 = -1
t2 = (-5 - 3) / 2 = -4
Step 3: Substitute the values of t back into x^2.
Now, we substitute these values of t back into the equation x^2 = t to find the values of x:
For t1 = -1, we have x^2 = -1. Taking the square root of both sides, we get:
x = ± √(-1)
Since the square root of -1 is denoted as i (the imaginary unit), we have:
x = ± i
For t2 = -4, we have x^2 = -4. Taking the square root of both sides, we get:
x = ± √(-4) = ± 2i
Therefore, the complex zeros of the polynomial f(x) = x^4 + 5x^2 + 4 are:
x1 = i
x2 = -i
x3 = 2i
x4 = -2i