Please help!! I do not understand any of this!!
Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4; zeros: 2, multiplicity 2; 3i
Iam not certain what you mean by multiplicity.
I see the roots as (x-2)(x-2)(x-3i)(x+3i)
that is degree four,multiple roots at 2, root at 3i, and -3i, and all real coefficients.
I'd be happy to help you with that! Let's break down the process step-by-step.
To form a polynomial, we need to start with the zero(s) given. In this case, we have two zeros: 2 with a multiplicity of 2, and 3i.
For a zero of 2 with multiplicity 2, it means that the factor (x - 2) will occur twice in the polynomial. So, we can start by writing:
f(x) = (x - 2)(x - 2)
Next, we have the zero 3i. Since this is a complex number, its conjugate (-3i) will also be a zero. Therefore, we can include both zeros by multiplying the polynomial with the factors (x - 3i) and (x + 3i):
f(x) = (x - 2)(x - 2)(x - 3i)(x + 3i)
Now, we can simplify this polynomial. First, let's combine the factors with the same expression in the parentheses:
f(x) = (x - 2)^2(x^2 - (3i)^2)
To simplify further, we'll need to calculate (3i)^2. The square of the imaginary unit i is -1, so:
(3i)^2 = 3^2 * i^2 = 9 * -1 = -9
Now, we can substitute this result back into the polynomial:
f(x) = (x - 2)^2(x^2 + 9)
At this point, we have formed a polynomial of degree 4 with real coefficients that has the given zeros.
I hope this explanation helps you understand the process! Let me know if there's anything else I can assist you with.