The roots of the cubic polynomial f ( x ) = − 2 x 3 + 5 x 2 + 4 x − 3 are x = − 1 , 0.5 , 3 . These values satisfy the equation f ( x ) = 0 . Verifying each root shows they indeed yield zero when substituted back into the function.
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Find the roots of the cubic function f ( x ) = − 2 x 3 + 5 x 2 + 4 x − 3 = 0 .
Use a tool to identify potential roots: x = − 1 , 0.5 , 3 .
Verify each root by substituting it back into the original equation.
The solutions are − 1 , 0.5 , 3 .
Explanation
Understanding the Problem We are given the cubic function f ( x ) = − 2 x 3 + 5 x 2 + 4 x − 3 and asked to solve for x when f ( x ) = 0 . This means we need to find the roots of the equation − 2 x 3 + 5 x 2 + 4 x − 3 = 0 .
Finding Possible Roots We can use the Rational Root Theorem to find possible rational roots. The possible rational roots are ± 1 , ± 3 , ± 2 1 , ± 2 3 . We can test these roots by substituting them into f ( x ) . Alternatively, we can use a tool to find the roots directly.
Verifying the Roots Using a tool, we find the roots to be x = − 1 , x = 0.5 , x = 3 . We can verify these roots by plugging them back into the original equation:
Checking x = -1 For x = − 1 : f ( − 1 ) = − 2 ( − 1 ) 3 + 5 ( − 1 ) 2 + 4 ( − 1 ) − 3 = − 2 ( − 1 ) + 5 ( 1 ) − 4 − 3 = 2 + 5 − 4 − 3 = 0
Checking x = 0.5 For x = 0.5 : f ( 0.5 ) = − 2 ( 0.5 ) 3 + 5 ( 0.5 ) 2 + 4 ( 0.5 ) − 3 = − 2 ( 0.125 ) + 5 ( 0.25 ) + 2 − 3 = − 0.25 + 1.25 + 2 − 3 = 0
Checking x = 3 For x = 3 : f ( 3 ) = − 2 ( 3 ) 3 + 5 ( 3 ) 2 + 4 ( 3 ) − 3 = − 2 ( 27 ) + 5 ( 9 ) + 12 − 3 = − 54 + 45 + 12 − 3 = 0
Final Answer Since all three values satisfy the equation f ( x ) = 0 , the solutions are x = − 1 , 0.5 , 3 .
Examples
Cubic functions can model various real-world phenomena, such as the volume of a container as a function of its dimensions, or the trajectory of a projectile. Solving for the roots of a cubic function helps us find critical points, such as the dimensions that maximize the volume or the launch angle that achieves a certain range. Understanding the roots of polynomial functions is crucial in engineering, physics, and economics for optimizing designs and predicting outcomes.