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I'm not sure how synthetic division would help here either. The usual way to do this is to show that the function is either increasing or decreasing on the interval, and the values at the endpoints have opposite signs. Then by the Intermediate Value Theorem (the function is continuous so we can use this), there will be exactly one zero in the interval.
The function `f(x)=x^4+3x+1` is decreasing on the interval `[-2,-1].` To show this, we take the derivative.
`f'(x)=4x^3+3,` which is less than zero whenever
`4x^3<-3,` which is the same as`x^3<-3/4,` or equivalently
`x<(-3/4)^(1/3).` Since `(-3/4)^(1/3)` is greater than `-1,` the derivative is negative on the whole interval, which means the function is decreasing and thus can have at most one zero there.
Now, `f(-2)=(-2)^4+3(-2)+1=11,` which is greater than 0, and `f(-1)=(-1)^4+3(-1)+1=-1,` less than 0.
Combining all of this, we can say that the graph of `f(x)=x^4+3x+1` crosses the `x` axis exactly once on the interval `[-2,-1],` and so has exactly one zero there.
Here's the graph to illustrate the theorems.
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