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Question 1177172: https://gyazo.com/1bd631884bbd97b5df97252bfbf0fd6d
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

The question:

Given the polynomial function a(n)x^n+a(n-1)x^(n-1)+...+a(1)x+a(0), what do we know about the number of roots of the polynomial if a(n)<0 and a(0)>0?

a(n)<0 tells us that for sufficiently large negative values of x the function value will be negative.

a(0)>0 tells us that f(0) is positive.

Together the two pieces of information tell us that there is at least one negative root.

There is nothing in the given information that tells us anything more about the number of roots.

ANSWER: a(n)<0 and a(0)>0 tells us that the function has at least one root.

Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website!
.

                I have another solution and different answer.

We are given a polynomial of EVEN degree f(x) = a(n)x^n + a(n-1)x^(n-1) + . . . + a(1)x + a(0) with negative leading coefficient a(n) < 0 and with positive constant term a(0) > 0. It means that the polynomial f(x) is negative and remains to be negative at x ---> -oo and also is negative and remains to be negative at x ---> oo. At the same time, the polynomial f(x) has positive value at x= 0: f(0) = a(0) > 0 (given). Hence, under given/imposed conditions, the polynomial f(x) has at least two zeroes: - at lest one positive root and - at least one negative root. ANSWER. At given conditions, the polynomial f(x) has at least two x-intercepts: at least one positive x-intercept and at least one negative x-intercept.

Solved, answered and explained. And completed.

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As a visual model and as a visual check, imagine a quadratic polynomial y = -x^2 + 1.