SOLUTION: If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots? I first attempted this problem by using Descartes' Rule of signs, and got 2 positive roots. When going to do the

Algebra ->  Trigonometry-basics -> SOLUTION: If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots? I first attempted this problem by using Descartes' Rule of signs, and got 2 positive roots. When going to do the      Log On


   

Question 888650: If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots?
I first attempted this problem by using Descartes' Rule of signs, and got 2 positive roots. When going to do the negative roots I got lost. I now think maybe this problem is not Descartes Rule of signs, but I am unsure.

Thank you so much for you time am help.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Why are you using Descartes' rule of signs? You'd have to expand the whole thing first.

f(x) = 0 --> 2x-1 = 0 or x^2 + 1 = 0 or (x-5)^2 = 0, so we have x = 1/2 or x = 5 (double root). x^2 + 1 = 0 gives no real roots.