Form a fifth-degree polynomial function with real coefficients such that [tex]$i, 1-2 i$[/tex], and -5 are zeros and [tex]$f(0)=75$[/tex].

[tex]$f(x)= \square$[/tex]
(Simplify your answer. Type an expression using [tex]$x$[/tex] as the variable.)

Identify the roots of the polynomial, including complex conjugates: i , − i , 1 − 2 i , 1 + 2 i , − 5 .
Write the polynomial in factored form: f ( x ) = a ( x − i ) ( x + i ) ( x − ( 1 − 2 i )) ( x − ( 1 + 2 i )) ( x + 5 ) .
Simplify the factors and use the condition f ( 0 ) = 75 to find the leading coefficient a = 3 .
Expand the polynomial to obtain the final expression: f ( x ) = 3 x 5 + 9 x 4 − 12 x 3 + 84 x 2 − 15 x + 75 .

f ( x ) = 3 x 5 + 9 x 4 − 12 x 3 + 84 x 2 − 15 x + 75 ​
Explanation

Identify the roots We are given that the polynomial has real coefficients, degree 5, and zeros i , 1 − 2 i , − 5 . Since the polynomial has real coefficients, the complex conjugate of each complex root is also a root. Thus, − i and 1 + 2 i are also roots. We also know that f ( 0 ) = 75 .

Write the polynomial in factored form Since the roots are i , − i , 1 − 2 i , 1 + 2 i , − 5 , the polynomial can be written as f ( x ) = a ( x − i ) ( x + i ) ( x − ( 1 − 2 i )) ( x − ( 1 + 2 i )) ( x + 5 ) , where a is a real constant.

Simplify the factors Simplify the factors: ( x − i ) ( x + i ) = x 2 − i 2 = x 2 + 1 and ( x − ( 1 − 2 i )) ( x − ( 1 + 2 i )) = ( x − 1 + 2 i ) ( x − 1 − 2 i ) = ( x − 1 ) 2 − ( 2 i ) 2 = x 2 − 2 x + 1 + 4 = x 2 − 2 x + 5 . Thus, f ( x ) = a ( x 2 + 1 ) ( x 2 − 2 x + 5 ) ( x + 5 ) .

Find the leading coefficient Use the condition f ( 0 ) = 75 to find the value of a . Substitute x = 0 into the polynomial: f ( 0 ) = a ( 0 2 + 1 ) ( 0 2 − 2 ( 0 ) + 5 ) ( 0 + 5 ) = a ( 1 ) ( 5 ) ( 5 ) = 25 a . Since f ( 0 ) = 75 , we have 25 a = 75 , so a = 3 .

Write the polynomial with the leading coefficient Therefore, f ( x ) = 3 ( x 2 + 1 ) ( x 2 − 2 x + 5 ) ( x + 5 ) .

Expand the polynomial Expand the polynomial to get the final expression for f ( x ) . First, expand ( x 2 + 1 ) ( x 2 − 2 x + 5 ) = x 4 − 2 x 3 + 5 x 2 + x 2 − 2 x + 5 = x 4 − 2 x 3 + 6 x 2 − 2 x + 5 . Then, multiply by ( x + 5 ) : ( x 4 − 2 x 3 + 6 x 2 − 2 x + 5 ) ( x + 5 ) = x 5 − 2 x 4 + 6 x 3 − 2 x 2 + 5 x + 5 x 4 − 10 x 3 + 30 x 2 − 10 x + 25 = x 5 + 3 x 4 − 4 x 3 + 28 x 2 − 5 x + 25 . Finally, multiply by 3: f ( x ) = 3 ( x 5 + 3 x 4 − 4 x 3 + 28 x 2 − 5 x + 25 ) = 3 x 5 + 9 x 4 − 12 x 3 + 84 x 2 − 15 x + 75 .


Examples
Polynomial functions are used to model various real-world phenomena, such as the trajectory of a projectile, the growth of a population, or the behavior of electrical circuits. In this case, we constructed a fifth-degree polynomial function with specific zeros and a given value at x = 0 . This type of problem can be applied in engineering to design systems with desired characteristics or in physics to model complex systems with multiple interacting components. For example, if you are designing a filter for an audio signal, you might use a polynomial to represent the frequency response of the filter. The roots of the polynomial would correspond to the frequencies that the filter blocks, and the value of the polynomial at x = 0 would correspond to the gain of the filter at DC.

Answered by GinnyAnswer | 2025-07-05

To create a fifth-degree polynomial with given zeros and real coefficients, we identified the roots, wrote the polynomial in factored form, and simplified. We found the leading coefficient using the condition f ( 0 ) = 75 , and then expanded the polynomial to arrive at the final expression. The resulting polynomial is f ( x ) = 3 x 5 + 9 x 4 − 12 x 3 + 84 x 2 − 15 x + 75 .
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Answered by Anonymous | 2025-07-07