Polynomial equation

Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

n=3

-2, 9+2i are zeros

f(1)=204

Answer 1

There are a few critical things to know to solve this problem:

1) a degree 3 polynomial has 3 zeros

You were given 2 so you need the other one. Whenever there is an complex zero (real + imaginary), the conjugate is also always a zero. The conjugate has the same numbers but opposite sign, so the conjugate of "9 + 2i" is "9 - 2i"

2) The zeros of a polynomial can be made into factors by prefacing them with "x - "

So the factors of this polynomial are [x - (-2)], [x - (9 + 2i)], and [[x - (9 - 2i)] or (x + 2), (x - 9 - 2i), and (x - 9 + 2i)

3) We can get a polynomial by multiplying the factors

So, in this case, we can multiply (x + 2)(x - 9 - 2i)(x - 9 + 2i)

Multiply (x - 9 - 2i)(x - 9 + 2i) like this:

So the total is x² - 18x + 85

Now multiply by (x + 2) the same way.

4) Realize that there are many possible polynomials because (x + 2) (x - 9 - 2i)(x - 9 + 2i) has the same zeros as 2(x + 2)(x - 9 - 2i)(x - 9 + 2i) which has the same zeros as 3(x + 2) (x - 9 - 2i)(x - 9 + 2i), etc.

So to find the specific polynomial in question, we need to include a multiplier variable:

f(x) = a(x + 2)(x - 9 - 2i)(x - 9 + 2i)

We then plug in x = 1 and f(x) = 204 and solve for "a"

Answer 2

P(x) = a(x + 2)(x - (9 + 2i))(x - (9 - 2i))

P(x) = a(x + 2)(x² - 18x + 85)

P(x) = a(x³ - 16x² + 49x + 170)

1 = a - 16a + 49a + 170

Can you solve for a and answer?