Question 204290

First count the sign changes of {{{f(a)=a^5-4a^2-7}}}


From {{{a^5}}} to {{{-4a^2}}}, there is a sign change from positive to negative 


From {{{-4a^2}}} to {{{-7}}}, there is no change in sign


So there is 1 sign change for the expression {{{f(a)=a^5-4a^2-7}}}. 


So there is 1 positive real zero




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{{{f(-a)=(-a)^5-4(-a)^2-7}}} Now let's replace each {{{a}}} with {{{-a}}}



{{{f(-a)=-a^5-4a^2-7}}} Simplify. Note: only the terms with odd exponents will have a sign change.



Now let's count the sign changes of {{{f(-a)=-a^5-4a^2-7}}}


From {{{-a^5}}} to {{{-4a^2}}}, there is no change in sign


From {{{-4a^2}}} to {{{-7}}}, there is no change in sign


So there are no sign changes for the expression {{{f(-a)=-a^5-4a^2-7}}}



So there are 0 negative real zeros



Note: if you graph {{{f(a)=a^5-4a^2-7}}}, you will find that there is indeed one positive real zero.