SOLUTION: Given P(x)=2x^5 + 7x^4 - 2x^3 - 25x^2 - 12x + 12 2. Using the DeCartes' rule of signs, state how many positive and how many negative real roots are possible for P(x).(Remember

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Given P(x)=2x^5 + 7x^4 - 2x^3 - 25x^2 - 12x + 12 2. Using the DeCartes' rule of signs, state how many positive and how many negative real roots are possible for P(x).(Remember      Log On


   



Question 131244: Given P(x)=2x^5 + 7x^4 - 2x^3 - 25x^2 - 12x + 12

2. Using the DeCartes' rule of signs, state how many positive and how many negative real roots are possible for P(x).(Remember that rational roots are real!)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Using Descartes' Rule of Signs, we can find the possible number of positive roots (x-intercepts that are positive) and negative roots (x-intercepts that are negative)

First lets find the number of possible positive real roots:

For 2x%5E5+%2B+7x%5E4+-+2x%5E3+-+25x%5E2+-+12x+%2B+12, simply count the sign changes

Here is the list of sign changes:
  1. 7x%5E4+ to +-+2x%5E3++ (positive to negative)
  2. +-+12x+ to 12+ (negative to positive)



So there are 2 sign changes, this means there are a maximum of 2 positive roots

So there could be 2, or 0 positive real roots







Now to find the number of negative real roots, we need to find f%28-x%29


f%28x%29=2x%5E5+%2B+7x%5E4+-+2x%5E3+-+25x%5E2+-+12x+%2B+12


Replace each x with -x


f%28-x%29=-2x%5E5+%2B+7x%5E4+%2B2x%5E3+-+25x%5E2+%2B+12x+%2B+12 Simplify


Now let's count the sign changes for -2x%5E5+%2B+7x%5E4+%2B2x%5E3+-+25x%5E2+%2B+12x+%2B+12




Here is the list of sign changes:
  1. -2x%5E5+ to +7x%5E4++ (negative to positive)
  2. 2x%5E3+ to -+25x%5E2+ (positive to negative)
  3. -+25x%5E2+ to 12x+ (negative to positive)




So there are 3 sign changes, this means there are a maximum of 3 negative roots

So there could be 3, or 1 negative real roots