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Bezzdna [24]
1 year ago
8

Find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utili

ty, use it to graph the function and verifythe real zeros and the given function value.n = 3;2 and 2 i are zeros;f(-1)=30f(x) =(Type an expression using x as the variable. Simplify your answer.)Help me solve this►View an example Get more helpClear allCheck answerCopyright © 2022 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy | Permissions | Contact UCon
Mathematics
1 answer:
Marrrta [24]1 year ago
6 0

We need to find a polynomial function f(x) with degree 3 and zeros 2 and 2i, such that:

f(-1)=30

Since 2i is a zero, -2i is also a zero of this function. Thus, we have:

f(x)=a(x-2)(x-2\imaginaryI)(x+2\imaginaryI)

where a is a constant.

Expanding the expression on the right side, we obtain:

\begin{gathered}  \\ f(x)=a(x-2)\lbrack x²-(2i)²\rbrack \\  \\ f(x)=a(x-2)(x²-4i²),\text{ i^^b2=-1} \\  \\ f(x)=a(x-2)(x²+4) \\  \\ f(x)=a(x³-2x²+4x-8) \end{gathered}

Now, using f(-1) = 30, we obtain:

\begin{gathered} 30=a\lbrack(-1)³-2(-1)²+4(-1)-8\rbrack \\  \\ 30=a(-1-2-4-8) \\  \\ 30=a(-15) \\  \\ a=\frac{30}{-15} \\  \\ a=-2 \end{gathered}

Therefore, the function is:

Answer

f(x)=-2x^{3}+4x^{2}-8x+16

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