Question

Show that each polynomial function has a real zero as described in parts (a) and (b). In Exercises 31 and 32, also work part (c). ƒ(x)=$$6x^4+13x^3$-11x^2-3x+5 no $$zero greater than 1

Accepted Answer

First, understand that the problem asks to show that the polynomial function $$f(x) = 6x^4$ + 13x^3$ - 11x^2$ - 3x + 5$$ has a real zero, and specifically that there is no zero greater than 1. Evaluate the polynomial at $$x = 1 by $$substituting 1 into the function: calculate $$f(1) = 6(1)^4$ + 13(1)^3$ - 11(1)^2$ - 3(1) + 5 to $$determine the sign of the function at 1. Evaluate the polynomial at another value less than 1, for example at $$x = 0$$, by calculating $$f(0) = 6(0)^4$ + 13(0)^3$ - 11(0)^2$ - 3(0) + 5 to $$check the sign of the function at 0. Use the Intermediate Value Theorem, which states that if a continuous function changes sign over an interval, then it must have at least one zero in that interval. Since polynomials are continuous everywhere, if $$f(0)$$ and $$f(1)$$ have opposite signs, there is at least one zero between 0 and 1. To show there is no zero greater than 1, analyze the behavior of $$f(x)$$ for $$x \u003e 1. $$You can check the sign of $$f(2) or $$consider the end behavior of the polynomial (leading term $$6x^4$$ dominates for large $$x$$), which tends to positive infinity, and verify that $$f(x)$$ does not cross zero for $$x \u003e 1.$$

Show that each polynomial function has a real zero as described in parts (a) and (b). In Exercises 31 and 32, also work part (c). ƒ(x)=6x^4+13x^3-11x^2-3x+5 no zero greater than 1

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Key Concepts

Polynomial Functions and Their Zeros

Bounds on Zeros of Polynomials

Synthetic Division