Find a polynomial function [tex]f(x)[/tex] of least degree having only real coefficients and the given zeros. Assume multiplicity 1 unless otherwise stated.

- Zero of 3 (multiplicity 2)
- Zero of [tex]4i[/tex]

[tex]f(x) = x^3 - 8x^2 + 22x - 20[/tex]

(Simplify your answer.)



Answer :

To find a polynomial function \( f(x) \) of the least degree with the given zeros, follow these steps:

1. Identify the given zeros:
- Zero at \( x = 3 \) with multiplicity 2.
- Zero at \( x = 4i \). Since the polynomial must have real coefficients, the complex conjugate \( x = -4i \) is also a zero.

2. Write the factors corresponding to each zero:
- For the zero at \( x = 3 \) with multiplicity 2: \((x - 3)^2\).
- For the zero at \( x = 4i \): \((x - 4i)\).
- For the zero at \( x = -4i \): \((x + 4i)\).

3. Combine all factors to form the polynomial:
Combine the factors to construct the polynomial:
[tex]\[ f(x) = (x - 3)^2 (x - 4i) (x + 4i) \][/tex]

4. Simplify the polynomial:
- First expand the complex conjugate pair product:
[tex]\[ (x - 4i)(x + 4i) = x^2 - (4i)^2 = x^2 - (-16) = x^2 + 16 \][/tex]
- The polynomial now is:
[tex]\[ f(x) = (x - 3)^2 (x^2 + 16) \][/tex]

5. Expand the remaining factor:
- Expand \((x - 3)^2\):
[tex]\[ (x - 3)^2 = x^2 - 6x + 9 \][/tex]
- Now multiply this with \(x^2 + 16\):
[tex]\[ f(x) = (x^2 - 6x + 9)(x^2 + 16) \][/tex]

6. Perform the final expansion:
- Multiply each term in the first polynomial by each term in the second polynomial:
[tex]\[ \begin{align*} f(x) &= x^2(x^2 + 16) - 6x(x^2 + 16) + 9(x^2 + 16) \\ &= x^4 + 16x^2 - 6x^3 - 96x + 9x^2 + 144 \\ &= x^4 - 6x^3 + 25x^2 - 96x + 144 \end{align*} \][/tex]

So, the polynomial of the least degree with the given properties is:
[tex]\[ f(x) = x^4 - 6x^3 + 25x^2 - 96x + 144 \][/tex]