help get the answer

Question

Find a polynomial function​ f(x) of least degree having only real coefficients and zeros as given. Assume multiplicity 1 unless otherwise stated.

zero of 3​ (multiplicity 2) and zero

3i

Answer 1

If 3 is a root, then (x - 3) is a factor of f(x). Since its multiplicity is given as 2, f(x) is starting to look like this:

f(x) = (x - 3)².

It is given that the imaginary number 3i is also a root. That means - 3i is also a root. That in turn means that there are additional factors of f(x) which are: (x - 3i) and [x - (-3i)] which is (x + 3i).

So f(x) looks like this in factored form:

f(x) = (x - 3)²(x - 3i) (x + 3i)

If you choose to multiply the out , you can square the binomial and find the product of (x - 3i) (x + 3i).

(x - 3)² = x² - 6x + 9

(x - 3i)(x + 3i) = x² - (3i)²= x²- 9i² = x² + 9

f(x ) = x⁴ - 6x³ + 9x² + 9x² - 54x + 81 = x⁴ - 6x³ + 18x² - 54x + 81

Visit the Desmos graph to enable functions where the root at x = 3 has different multiplicities.

desmos.com/calculator/ae55jftqhn