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## anonymous one year ago According to Descartes' rule of signs, how many possible negative real roots could the following polynomial function have?

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1. anonymous

2. anonymous

I think the answer is 2 or 0

3. anonymous

look at our function as $$x\to-\infty$$; this is made easier by substituting $$x\to -x$$ to get $$f(-x)=3x^4+5x^3+5x^2-5x+2$$. for small $$x$$, note that first $$2$$ will dominate and we will begin above the $$x$$ axis. then as $$x$$ grows, $$-5x$$ will dominate and could potentially drive our function below the $$x$$ axis -- this would necessitate the existence of a root.

4. anonymous

then as x continues to grow $$5x^2$$ comes to dominate, which might drive our function back above the $$x$$ axis and could potentially result in another root if our function was driven below the $$x$$ axis by the $$-5x$$ term -- or, in the case $$-5x$$ failed to pull our function below the $$x$$ axis, then neither root would exist. the rest of our terms as they dominate will only further pull our function higher and higher and risk no chance of introducing any new roots

5. anonymous

so we have two possibilities for such negative real roots $$-x$$: either $$-5x$$ was sufficiently 'strong' when it came to dominate that it drove our function below the $$x$$-axis only for the subsequent terms to eventually dominate and drive our function back above the $$x$$-axis to yield two such roots, or it failed to drive it below the $$x$$-axis at all and we have zero such roots

6. anonymous

this is Descartes' rule of signs explained intuitively -- we can count the number of possible 'crossings' of the $$x$$-axis each time a term of opposite sign comes to dominate; here, we have $$2 \to -5x$$ represent a possible crossing, followed by $$-5x\to 5x^2$$ representing another

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