algebra 2

Guest:

What are the roots for P(x) = x^3 + 2x^2 -4

Are you sure the signs are right? The answer is not popping out at me!

Hint: You have to look fo an x value that will make P(x)=0

then look here
http://www.mathsisfun.com/algebra/polynomials-remainder-factor.html

At least one root is between 1 and 2

According to Descartes rule of signs, we have one positive root, and either two negative roots or no negative roots

Let's see...I'm going to attempt this for the first time!!

Wow!! Those were pretty complicated!!

As suspected, only the third root is real (I'll let you calculate (THAT one!!) and the other two are complex.

Note.....if we have any complex roots in any polynomial, they ALWAYS occur in pairs. We can't have an odd number of complex roots.

Guest:

What are the roots for P(x) = x^3 + 2x^2 -4

Quote:

At least one root is between 1 and 2

According to Descartes rule of signs, we have one positive root, and either two negative roots or no negative roots

Okay Chris,
I am not familiar with Descartes rule - Yes I know I could look it up - okay I did.

Now P(1) is negative and P(2) is positive therefore there must be at least root between 1 and 2. That is obvious.

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs
EXTRACT
"Positive roots[edit]
The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

Are 2x^2 -4 counted as consecutive since there is a zero term inbetween ? If they are then there is 1 sign change. Hence there is 1 positive root.

Could you comment please Chris?
Thankyou.

PS i still think it was a copy error.