Question 245808

First count the sign changes of {{{f(x)=-6x^5+x^4+5x^3+x+1}}}


From {{{-6x^5}}} to {{{x^4}}}, there is a sign change from negative to positive 


From {{{x^4}}} to {{{5x^3}}}, there is no change in sign


From {{{5x^3}}} to {{{x}}}, there is no change in sign


From {{{x}}} to {{{1}}}, there is no change in sign


So there is 1 sign change for the expression {{{f(x)=-6x^5+x^4+5x^3+x+1}}}. 


So there is 1 positive zero.




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{{{f(-x)=-6(-x)^5+(-x)^4+5(-x)^3+(-x)+1}}} Now let's replace each {{{x}}} with {{{-x}}}



{{{f(-x)=6x^5+x^4-5x^3-x+1}}} Simplify



Now let's count the sign changes of {{{f(-x)=6x^5+x^4-5x^3-x+1}}}


From {{{6x^5}}} to {{{x^4}}}, there is no change in sign


From {{{x^4}}} to {{{-5x^3}}}, there is a sign change from positive to negative 


From {{{-5x^3}}} to {{{-x}}}, there is no change in sign


From {{{-x}}} to {{{1}}}, there is a sign change from negative to positive 


So there are 2 sign changes for the expression {{{f(-x)=6x^5+x^4-5x^3-x+1}}}. 


So there are 2 or 0 negative zeros