SOLUTION: Is this right; according to Descartes' rule of signs how many negative roots, and how many positive real roots are there in this problem; f(x) =3x^3+9x^2+8x so there are 3 positive

Algebra ->  Functions -> SOLUTION: Is this right; according to Descartes' rule of signs how many negative roots, and how many positive real roots are there in this problem; f(x) =3x^3+9x^2+8x so there are 3 positive      Log On


   

Question 204288: Is this right; according to Descartes' rule of signs how many negative roots, and how many positive real roots are there in this problem; f(x) =3x^3+9x^2+8x so there are 3 positive real roots and 1 negative root??
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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%28x%29=3x%5E3%2B9x%5E2%2B8x

From 3x%5E3 to 9x%5E2, there is no change in sign

From 9x%5E2 to 8x, there is no change in sign

So there are no sign changes for the function f%28x%29=3x%5E3%2B9x%5E2%2B8x


So there are 0 positive zeros



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f%28-x%29=3%28-x%29%5E3%2B9%28-x%29%5E2%2B8%28-x%29 Now let's replace each x with -x


f%28-x%29=-3x%5E3%2B9x%5E2-8x Simplify. Note: only the terms with odd exponents will have a change in sign.


Now let's count the sign changes of f%28-x%29=-3x%5E3%2B9x%5E2-8x

From -3x%5E3 to 9x%5E2, there is a sign change from negative to positive

From 9x%5E2 to -8x, there is a sign change from positive to negative

So there are 2 sign changes for the function f%28-x%29=-3x%5E3%2B9x%5E2-8x.

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%28x%29=3x%5E3%2B9x%5E2%2B8x+, 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).