# Polynomial Analysis

## Homework Statement

Let f(x)=a0+a1x+a2x2+a3x3+a4x4. Show that if a0a4<0, then f(x)=0 have at least 2 real solutions.

## The Attempt at a Solution

Any hints? I can't tell how to begin an attempt for a solution.

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Office_Shredder
Staff Emeritus
Gold Member
a0a4<0 so the two coefficients are of opposite sign. Let's assume a0>0 for a second. Can you tell me where f(x) is positive? Also, two different places f(x) will be negative

Well if a0>0, then in order for f(x) to be positive, a1x+a2x2+a3x3+a4x4<a0. What does this tell me about anything?

Office_Shredder
Staff Emeritus
Gold Member
That's not the requirement for f(x) to be positive, but that's OK. There's an easy to find value of x that makes f(x) positive. Think about how you would try to graph f(x) and you should see it

How can I find the value of x when I only know the signs of a0 and a4 and nothing about a2 and a3?

Office_Shredder
Staff Emeritus