Polynomial Analysis: Show 2 Real Solutions for f(x)=0

[toddlinsley79]

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.

Homework Equations

The Attempt at a Solution

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

 

[Office_Shredder]

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

 

[toddlinsley79]

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]

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

 

[toddlinsley79]

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]

For what value of x does f(x) only depend on a₀?