Find all real and non-real roots of the function [tex]$f(x)=\left(x^2+4\right)(x-2)$[/tex].
A) [tex]$x= \pm 2 i$[/tex]
B) [tex]$x= \pm 4 i, 2$[/tex]
C) [tex]$x= \pm 2$[/tex]
D) [tex]$y= \pm 2 i, 2$[/tex]
Answer (1)
To find the roots of f ( x ) = ( x 2 + 4 ) ( x − 2 ) :
Set f ( x ) = 0 , leading to ( x 2 + 4 ) ( x − 2 ) = 0 .
Solve x 2 + 4 = 0 to find non-real roots x = ± 2 i .
Solve x − 2 = 0 to find the real root x = 2 .
The roots are x = ± 2 i , 2 .
Explanation
Understanding the Problem We are asked to find all real and non-real roots of the function f ( x ) = ( x 2 + 4 ) ( x − 2 ) . This means we need to find all values of x for which f ( x ) = 0 .
Setting up the Equation To find the roots of the function, we set f ( x ) = 0 and solve for x :
( x 2 + 4 ) ( x − 2 ) = 0
Solving for the Roots This equation is satisfied if either x 2 + 4 = 0 or x − 2 = 0 . Let's solve each of these equations separately.
First, consider x 2 + 4 = 0 . Subtracting 4 from both sides gives x 2 = − 4 . Taking the square root of both sides gives x = ± − 4 = ± 2 i . So the non-real roots are x = 2 i and x = − 2 i .
Next, consider x − 2 = 0 . Adding 2 to both sides gives x = 2 . So the real root is x = 2 .
Final Answer Therefore, the roots of the function f ( x ) = ( x 2 + 4 ) ( x − 2 ) are x = 2 i , − 2 i , and 2 .
Examples
Finding the roots of a polynomial function is a fundamental concept in algebra and calculus. For instance, engineers use roots of functions to determine the stability of systems, such as bridges or electrical circuits. By finding the values at which a function equals zero, they can identify critical points that may indicate potential failure or instability. This ensures designs are safe and reliable.
