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.
3x^3+5x^2-4x+3=0
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
-3x^3+5x^2+4x+3=0
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.
x^2-2x-4=0
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:
