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Ch. 3 - Polynomial and Rational Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063당신이 사용하는 게 아니라요?교과서 변경
4장, 문제 45

For each polynomial function, use the remainder theorem to find ƒ(k). ƒ(x) = 6x4 + x3 - 8x2 + 5x+6; k=1/2

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Recall the Remainder Theorem: For a polynomial ƒ(x), the remainder when divided by (x - k) is equal to ƒ(k). So, to find ƒ(k), we simply evaluate the polynomial at x = k.
Write down the polynomial function and substitute x with k = \(\frac{1}{2}\): \[ƒ\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + 6\]
Calculate each term separately: - Compute \(6\left(\frac{1}{2}\right)^4\) - Compute \(\left(\frac{1}{2}\right)^3\) - Compute \(-8\left(\frac{1}{2}\right)^2\) - Compute \(5\left(\frac{1}{2}\right)\) - The constant term is 6
Add all the computed values together to find the value of ƒ\(\left\)(\(\frac{1}{2}\)\(\right\)). This sum represents the remainder when ƒ(x) is divided by \(x - \frac{1}{2}\).
Thus, the value of ƒ(k) is the remainder according to the Remainder Theorem, completing the problem.

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