|On 3/21/01 6:36:35 PM, Kayla Woodberry wrote:
I need help! We are studying Descartes' Rule of Signs and Bounds on Roots, and I do not understand this at all.
Here is an example question:
"Use Descartes' rule of signs to find the number of possible positive, negative, and nonreal roots of each equation. Do not try to find the roots."
"Tell how many roots each equation has."
Just two more......
"Write a third-degree polynomial equation with real coefficients and the given roots."
"Find integer bounds for the roots of each equation."
If someone could explain to me how to solve these things....maybe I could actually do my homework! Thanks in advance!
|Use Descartes' rule of signs
to find the number of possible positive, negative, and nonreal roots of
this equation. Do not try to find the roots.
Descartes' rule of signs says "Count up the number of changes of signs when the polynomial is in standard form. The number of positive real solutions will be that, or that minus an even number." This example changes sign twice, so the there are at most two positive real solutions.
To answer the same question about negative roots, note that the negative roots of f(x) are the positive roots of f(-x). To change polynomial f(x) into f(-x), change the signs of all the odd order terms. So f(-x) in your example is
This changes sign once, so the number of positive roots of f(-x) is one, so there is one negative root of f(x). Thus from Descartes' Rule of Signs, we can conclude there are either zero or two positive roots, either two or zero (respectively) nonreal roots, and exactly one negative root.
An nth order polynomial has n roots, so -32x^111-x^5=1 has 111 roots.
A 3rd degree polynomial with roots 2 and 2+i and real coefficients must also have 2-i as its third root, because the complex roots come in conjugate pairs. To find the polynomial, multiply (x-2)(x-2-i)(x-2+i). I'll leave that to you. Post another message if you need more help with this
Find integer bounds for the roots of the equation.
Test some roots using synthetic division. When you do this, you get P(x)=Q(x)(x-r)+R. If all the coefficients of your quotient Q(x) are positive, and x is positive, and R is positive, and x>r, then P(x) must be positive, so in particular it can't be 0. This means in this case r is an upper bound for the roots of P. In practice, you use this by doing the synthetic division, and looking at all the numbers at the bottom. If they are all positive, then you have an upper bound.
You can also get a negative number as a lower bound by doing the same thing with the corresponding positive number and P(-x). This works because the roots of P(-x) are all negatives of the roots of P(x). To find P(-x), you just reverse the signs of all the odd-order terms of P(x). Then, if the high-order sign is negative, you can find -P(-x), which has the same roots as P(-x), and thus the same negative lower bound.
Now, for your example...
Try various positive values of r to find a Q(x)(x-r)+R such that all the coefficients of Q(x) and R are all positive. You'll find before long that r=4 works for this equation:
So 4 is an upper bound of the positive roots of x^2-2x-4=0.
Now let's find a negative lower bound. This would be the negative of a positive upper bound of f(-x)=x^2+2x-4=0.
Again, using synthetic division, we quickly see that r=2 works:
So -2 is a lower bound of the negative roots of x^2-2x-4=0.
Cut the knot: Descartes’ Rule of Signs, by Scott E. Brodie, 1/1/99
Polynomial Remainder Theorem
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