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Shkiper50 [21]
3 years ago
6

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.

Mathematics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

\displaystyle f(x) = -2(x-2)(x^2+4)

Step-by-step explanation:

We want to find a third degree polynomial with zeros <em>x </em>= 2 and <em>x</em> = 2i and f(-1) = 30.

First, note that by the Complex Root Theorem, since <em>x</em> = 2i is a root, <em>x</em> = -2i must also be a root.

Hence, we will have the three factors:

\displaystyle f(x) = a(x-(2))(x-(2i))(x-(-2i))

Where <em>a</em> is the leading coefficient.

Expand and simplify the second and third factors:

\displaystyle \begin{aligned} (x-(2i))(x-(-2i)) &= (x-2i)(x+2i) \\ \\ &= x(x-2i)+2i(x-2i) \\ \\ &= (x^2 - 2ix) + (2ix - 4i^2) \\ \\ &=x^2 + 4\end{aligned}

Hence:

\displaystyle f(x) = a(x-2)(x^2+4)

Since f(-1) = 30:

\displaystyle \begin{aligned}  f(x) &= a(x-2)(x^2+4) \\ \\ (30) &= a((-1)-2)((-1)^2+4) \\ \\ 30 &= -15a \\ \\ a&= -2\end{aligned}

In conclusion, third degree polynomial function is:

\displaystyle f(x) = -2(x-2)(x^2+4)

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