Math III 4.6
"Your friend is using Descartes's Rule of Signs to find the number of negative real roots of x^3 + x^2 + x + 1 = 0. Describe and correct the error."
"P(-x) = (-x)^3 +(-x)^2 + (-x) +1
= -x^3 - x^2 - x +1"
"Because there is only one sign change in P(-x), there must be one negative real root."
- In Decartes's rule of sign, the number of negative real roots of p(x)=0 is either equal to the number of sign changers between consecutive coefficients of p(-x) or is less than that by an even number.
This problem is wrong because she changed p(x) to p(-x) when it should've stayed positive. When she switched the signs it made x^2 a -x^2. In addition to the error, it would've had three sign changes instead of one if the p(x) would have stayed a positive.
First you need to count the number of sign changes in the equation p(x) to find out the number of possibilities..
According to the rule, if there are 3 sign changes and if its odd.. you use 3 and count by odd numbers to zero.
Therefore, it is 3 or 1 negative real roots.
A gardener is designing a new garden in the shape of a trapezoid. She wants the shorter base to be twice the height and the longer base to be 4 feet longer than the shorter base. If she has enough topsoil to create a 60ft^2 garden, what dimensions should she use for the garden?
Plug in the information to the formula: