Form a fifth-degree polynomial function with real coefficients such that [tex]$2 i, 1-i$[/tex], and 1 are zeros and [tex]$f(0)=-24$[/tex].

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

Find the complex conjugate roots: Since the polynomial has real coefficients, the complex conjugates − 2 i and 1 + i are also roots.
Write the polynomial in factored form: f ( x ) = a ( x − 2 i ) ( x + 2 i ) ( x − ( 1 − i )) ( x − ( 1 + i )) ( x − 1 ) .
Simplify and use f ( 0 ) = − 24 to find a : After simplifying, f ( x ) = a ( x 2 + 4 ) ( x 2 − 2 x + 2 ) ( x − 1 ) , and using f ( 0 ) = − 24 , we find a = 3 .
Expand the polynomial: f ( x ) = 3 x 5 − 9 x 4 + 24 x 3 − 42 x 2 + 48 x − 24 . The final answer is 3 x 5 − 9 x 4 + 24 x 3 − 42 x 2 + 48 x − 24 ​ .

Explanation

Identify the roots We are given that the polynomial has real coefficients and has zeros 2 i , 1 − i , and 1. Since the polynomial has real coefficients, the complex conjugate of each complex root is also a root. Thus, − 2 i and 1 + i are also roots. Therefore, the polynomial has roots 2 i , − 2 i , 1 − i , 1 + i , 1 . We are also given that f ( 0 ) = − 24 .

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

Simplify ( x − 2 i ) ( x + 2 i ) Let's simplify the expression: ( x − 2 i ) ( x + 2 i ) = x 2 − ( 2 i ) 2 = x 2 − 4 i 2 = x 2 − 4 ( − 1 ) = x 2 + 4 .

Simplify ( x − ( 1 − i )) ( x − ( 1 + i )) Now, let's simplify the expression: ( x − ( 1 − i )) ( x − ( 1 + i )) = ( x − 1 + i ) ( x − 1 − i ) = (( x − 1 ) + i ) (( x − 1 ) − i ) = ( x − 1 ) 2 − i 2 = x 2 − 2 x + 1 − ( − 1 ) = x 2 − 2 x + 2 .

Use the condition f ( 0 ) = − 24 Thus, f ( x ) = a ( x 2 + 4 ) ( x 2 − 2 x + 2 ) ( x − 1 ) . Now we use the condition f ( 0 ) = − 24 to find the value of a . Substituting x = 0 into the expression for f ( x ) , we get: f ( 0 ) = a ( 0 2 + 4 ) ( 0 2 − 2 ( 0 ) + 2 ) ( 0 − 1 ) = a ( 4 ) ( 2 ) ( − 1 ) = − 8 a .

Solve for a We solve the equation − 8 a = − 24 for a : a = − 8 − 24 ​ = 3 . Therefore, the polynomial is f ( x ) = 3 ( x 2 + 4 ) ( x 2 − 2 x + 2 ) ( x − 1 ) .

Expand the polynomial Now, let's expand the polynomial to get the final expression for f ( x ) . First, we multiply ( x 2 + 4 ) ( x 2 − 2 x + 2 ) = x 4 − 2 x 3 + 2 x 2 + 4 x 2 − 8 x + 8 = x 4 − 2 x 3 + 6 x 2 − 8 x + 8 . Then we multiply by ( x − 1 ) to get ( x 4 − 2 x 3 + 6 x 2 − 8 x + 8 ) ( x − 1 ) = x 5 − 2 x 4 + 6 x 3 − 8 x 2 + 8 x − x 4 + 2 x 3 − 6 x 2 + 8 x − 8 = x 5 − 3 x 4 + 8 x 3 − 14 x 2 + 16 x − 8 . Finally, we multiply by 3 to get f ( x ) = 3 ( x 5 − 3 x 4 + 8 x 3 − 14 x 2 + 16 x − 8 ) = 3 x 5 − 9 x 4 + 24 x 3 − 42 x 2 + 48 x − 24 .

Final Answer Therefore, the fifth-degree polynomial function is f ( x ) = 3 x 5 − 9 x 4 + 24 x 3 − 42 x 2 + 48 x − 24 .


Examples
Polynomial functions are used to model various real-world phenomena. For instance, engineers use polynomials to design curves for roads and bridges. Economists use them to model cost and revenue functions, helping businesses predict profits. In physics, polynomials can describe the trajectory of projectiles or the shape of lenses. Understanding how to construct and manipulate polynomial functions is essential for solving problems in these fields.

Answered by GinnyAnswer | 2025-07-05