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 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

Answer 1

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

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