<PRE><B>Question 166504</B>
# 1


Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let&#39;s list the factors of -15 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm3, \pm5, \pm15]


Now let&#39;s list the factors of 3 (the first coefficient):


*[Tex \LARGE q=\pm1, \pm3]


Now let&#39;s divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{1}{3}, \frac{3}{1}, \frac{3}{3}, \frac{5}{1}, \frac{5}{3}, \frac{15}{1}, \frac{15}{3}, \frac{-1}{1}, \frac{-1}{3}, \frac{-3}{1}, \frac{-3}{3}, \frac{-5}{1}, \frac{-5}{3}, \frac{-15}{1}, \frac{-15}{3}]







Now simplify


These are all the distinct rational zeros of the function that could occur


*[Tex \LARGE  1, \frac{1}{3}, 3, 5, \frac{5}{3}, 15, -1, \frac{-1}{3}, -3, -5, \frac{-5}{3}, -15]




&lt;hr&gt;


# 2



Now let&#39;s use Descartes Rule of Signs to find the number of positive, negative,and imaginary zeros.



First count the sign changes of {{{f(x)=9x^3-7x^2+10x-4}}}


From {{{9x^3}}} to {{{-7x^2}}}, there is a sign change from positive to negative 


From {{{-7x^2}}} to {{{10x}}}, there is a sign change from negative to positive 


From {{{10x}}} to {{{-4}}}, there is a sign change from positive to negative 


So there are 3 sign changes for the expression {{{f(x)=9x^3-7x^2+10x-4}}}. 


So there are 3 or 1 positive zeros



So if there are 3 positive zeros, then there will be NO imaginary zeros (since there a total of 3 zeros)


Or...


If there is only 1 positive zero, then there will be 2 imaginary zeros.




------------------------------------------------


{{{f(-x)=9(-x)^3-7(-x)^2+10(-x)-4}}} Now let&#39;s replace each {{{x}}} with {{{-x}}}



{{{f(-x)=-9x^3-7x^2-10x-4}}} Simplify



Now let&#39;s count the sign changes of {{{f(-x)=-9x^3-7x^2-10x-4}}}


From {{{-9x^3}}} to {{{-7x^2}}}, there is no change in sign


From {{{-7x^2}}} to {{{-10x}}}, there is no change in sign


From {{{-10x}}} to {{{-4}}}, there is no change in sign


So there are no sign changes for the expression {{{f(-x)=-9x^3-7x^2-10x-4}}}



So there are 0 negative zeros



==========================================


Answer:


So here are the possibilities:


3 positive zeros, 0 negative zeros, and 0 imaginary zeros


Or...


1 positive zero, 0 negative zeros, and 2 imaginary zeros</PRE>