# form a polynomial f(x) with real coefficients having the given degree & zeros degree 4; zero -3 -3i ; -3 multiplicity 2 polynomial f(x) = a( )must show work

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### 1 Answer

Notice that the polynomial has a complex solution `x = -3-3i` , hence, the complex conjugate `x = -3+3i` is also a root of polynomial.

The problem provides the information that the solution `x= -3` has n order of multiplicity `n=2` .

You should remember that you may write the factored form of a polynomial of fourth degree such that:

`f(x) = a(x-x_1)(x-x_2)(x-x_3)(x-x_4)`

You need to substitute the given roots for `x_1,x_2,x_3,x_4` such that:

`f(x) = a(x - (-3-3i))(x - (-3+3i))(x - (-3))(x - (-3))`

`f(x) = a(x + 3 + 3i)(x + 3 - 3i)(x + 3)^2`

`f(x) = a((x + 3)^2- (3i)^2)(x + 3)^2`

Using the fact that `i^2 = -1` yields:

`f(x) = a((x + 3)^2 + 9)(x + 3)^2`

**Hence, evaluating the given polynomial, under the given conditions, yields `f(x) = a((x + 3)^2 + 9)(x + 3)^2` .**

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