Use synthetic division and the Remainder Theorem to find the indicated function value.
[tex]f(x)=7 x^4+10 x^3+6 x^2-5 x+53 ; f(2)[/tex]
Answer (2)
Set up synthetic division with the coefficients of the polynomial f ( x ) and the value x = 2 .
Perform synthetic division to find the remainder.
Apply the Remainder Theorem, which states that the remainder is equal to f ( 2 ) .
Conclude that f ( 2 ) = 259 .
259
Explanation
Understanding the Problem and Remainder Theorem We are given the polynomial function f ( x ) = 7 x 4 + 10 x 3 + 6 x 2 − 5 x + 53 and we are asked to find the value of f ( 2 ) using synthetic division and the Remainder Theorem. The Remainder Theorem states that if we divide a polynomial f ( x ) by ( x − a ) , then the remainder is f ( a ) . In this case, we want to find f ( 2 ) , so we will divide f ( x ) by ( x − 2 ) using synthetic division.
Setting up Synthetic Division To perform synthetic division, we write down the coefficients of the polynomial f ( x ) and the value x = 2 . The coefficients are 7 , 10 , 6 , − 5 , and 53 . We set up the synthetic division as follows:
2 | 7 10 6 -5 53
|________________
Performing Synthetic Division Now, we perform the synthetic division:
Bring down the first coefficient, which is 7.
2 | 7 10 6 -5 53
|________________
7
Multiply 2 by 7 and write the result (14) under the next coefficient (10). Add them: 10 + 14 = 24 .
2 | 7 10 6 -5 53
| 14
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7 24
Multiply 2 by 24 and write the result (48) under the next coefficient (6). Add them: 6 + 48 = 54 .
2 | 7 10 6 -5 53
| 14 48
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7 24 54
Multiply 2 by 54 and write the result (108) under the next coefficient (-5). Add them: − 5 + 108 = 103 .
2 | 7 10 6 -5 53
| 14 48 108
|________________
7 24 54 103
Multiply 2 by 103 and write the result (206) under the next coefficient (53). Add them: 53 + 206 = 259 .
2 | 7 10 6 -5 53
| 14 48 108 206
|________________
7 24 54 103 259
Applying the Remainder Theorem The last number in the bottom row is the remainder, which is 259. According to the Remainder Theorem, f ( 2 ) = 259 .
Final Answer Therefore, the value of the function f ( x ) at x = 2 is 259.
Examples
Evaluating polynomial functions is crucial in many engineering applications, such as designing filters in signal processing or modeling physical systems. For instance, when designing a digital filter, engineers use polynomials to represent the filter's transfer function. Evaluating the polynomial at specific frequencies helps determine the filter's response at those frequencies, ensuring it meets the required specifications. Similarly, in structural engineering, polynomials can model the deflection of a beam under load, and evaluating the polynomial at different points along the beam helps engineers understand the stress distribution and ensure the structure's safety. Let's say the deflection of a beam is given by d ( x ) = 0.001 x 4 − 0.02 x 3 + 0.1 x 2 , where x is the distance from one end. To find the deflection at x = 5 , we calculate d ( 5 ) = 0.001 ( 5 ) 4 − 0.02 ( 5 ) 3 + 0.1 ( 5 ) 2 = 0.001 ( 625 ) − 0.02 ( 125 ) + 0.1 ( 25 ) = 0.625 − 2.5 + 2.5 = 0.625 . This tells us the beam deflects by 0.625 units at that point.
Using synthetic division and the Remainder Theorem, we evaluated the polynomial f ( x ) = 7 x 4 + 10 x 3 + 6 x 2 − 5 x + 53 at x = 2 and found that f ( 2 ) = 259 .
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