Question 204288
Please show your work as to how you got your answers. I'm not sure how you got them and where you went wrong.



First count the sign changes of {{{f(x)=3x^3+9x^2+8x}}}


From {{{3x^3}}} to {{{9x^2}}}, there is no change in sign


From {{{9x^2}}} to {{{8x}}}, there is no change in sign


So there are no sign changes for the function {{{f(x)=3x^3+9x^2+8x}}}



So there are 0 positive zeros




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{{{f(-x)=3(-x)^3+9(-x)^2+8(-x)}}} Now let's replace each {{{x}}} with {{{-x}}}



{{{f(-x)=-3x^3+9x^2-8x}}} Simplify. Note: only the terms with odd exponents will have a change in sign.



Now let's count the sign changes of {{{f(-x)=-3x^3+9x^2-8x}}}


From {{{-3x^3}}} to {{{9x^2}}}, there is a sign change from negative to positive 


From {{{9x^2}}} to {{{-8x}}}, there is a sign change from positive to negative 


So there are 2 sign changes for the function {{{f(-x)=-3x^3+9x^2-8x}}}. 


So there are 2 or 0 negative zeros. Note: you count down by 2 (since complex zeros come in conjugate pairs).



Note: if you graph {{{f(x)=3x^3+9x^2+8x }}}, you'll see that there are no positive or negative real zeros. It turns out that {{{x=0}}} is the only zero (which is neither positive nor negative).