|
Question 61240: In the equation P(x) = 2x3 + 7x2 + x + 3 how many variations of sign are there, in the sense of Descartes’ rule of signs? What is the number of positive real zeros of P(x)?
Found 2 solutions by funmath, jai_kos:
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! In the equation P(x) = 2x^3 + 7x^2 + x + 3 how many variations of sign are there, in the sense of Descartes’ rule of signs? What is the number of positive real zeros of P(x)?
:
First note that a polynomial cannot have more real zeroes than its degree. The most zeros this polynomial can have is 3, because this is a third degree polynomial.
:
The Descartes' Rule of Signs States:
Let f denote a polynomial function written in standard form.
*The number of positive real zeroes of f either equals the number of variations in the sign of the nonzero coefficients of f(x) or else equals that number less an even integer.
*The number of negative real zeroes of f either equlas the number of variations in the sign of f(-x) or else equals that number less an even integer.
:
P(x) is always positive (++)(++)(++), the terms never change from + to - or - to +, therefore there are no positive zeros.
:
P(-x)=2(-x)^3+7(-x)^2+(-x)+3
P(-x)=-2x^3+7x^2-x+3
Has 3 variations of signs (-+),(+-),(-+), that means you can have 3 (or 3-2=1) negative zeroes.
:
Happy Calculating!!!
Answer by jai_kos(139) (Show Source):
You can put this solution on YOUR website! Given P(x) = 2x^3 + 7x^2 + x + 3
If you chech there are no changes of the negative to positive.
Hence there are no positive roots.
Now replace x by -x in equation(1), we get
P(-x) = 2(-x)^3 + 7(-x)^2 + (-x) +3
P(-x) = -2x^3 + 7x^2 - x + 3
We have 3 changes from positive to negative or negative to positive.
Hence there are 3, 1 negative roots.
Therefore there are no positive roots and 3 , 1 negative roots.
|
|
|
| |
