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Chapter 3: Problem 62

Explain how the Remainder Theorem can be used to find \(f(-6)\) if \(f(x)=x^{4}+7 x^{3}+8 x^{2}+11 x+5 .\) What advantage is there to using the Remainder Theorem in this situation rather than evaluating \(f(-6)\) directly?

Short Answer

Expert verified
The Remainder Theorem provides a more efficient way to compute \(f(-6)\) for a high-degree polynomial rather than evaluating \(f(-6)\) by substituting \(-6\) directly into \(f(x)\). It lets us drastically simplify the calculation process, saving time and reducing possible computational errors.

Step by step solution

01

Setup for Long Division

We're concerned with \(x - (-6)\), which simplifies to \(x + 6\), and we're going to use it to divide the polynomial provided. This can be achieved by using synthetic or long division method.
02

Applying Long Division

Set up the polynomial \(f(x) = x^{4} + 7x^{3} + 8x^{2} + 11x + 5\) for long division by \(x + 6\). Divide \(x^{4}\) by \(x\) to get \(x^{3}\), multiply \(x^{3}\) by \(x+6\) to get \(x^{4} + 6x^{3}\). Then subtract \(x^{4} + 6x^{3}\) from \(x^{4} + 7x^{3}\) to get \(x^{3}\). Drag down the 8x^{2} and repeat the process.
03

Find the Remainder

Continue with the long polynomial division until there's no term left to bring down. The remaining value will be the remainder, which should be our value for \(f(-6)\) according to the Remainder theorem.
04

Interpretation

The final remainder will give us the result for \(f(-6)\). This simplification wouldn't be possible without the Remainder Theorem, thus underscoring the advantage of the Remainder theorem over direct substitution when handling complex polynomials or large, especially negative, numerical values.

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