
Review of the Factor Theorem
Recall from last time, if P(x) is a polynomial, and
P(r) = 0
then the remainder
produced when P(x) is divided by x  r is 0. We can conclude that
r
is a root of P(x) if and only if the x  r divides P(x)
Example:
Given that 3 is a root of
x^{3}  x^{2}  8x + 6
Find the other two roots.
Solution:
We use synthetic division to divide by x  3. The
result is
x^{2} +2x  2
Next we use the quadratic formula to find
x = 1 +
or x = 1 

The Conjugate Root Theorem
The Conjugate Root Theorem

Suppose that P(x) has a root of the form
a +
b
where c is
not a perfect square, then
a  b
is also a root.
 Suppose that P(x) has a root of the form
a + bi
then
a  bi
is
also a root.


The Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra
Every polynomial has a root in the complex numbers, moreover if the polynomial
has degree n then the polynomial can be written as a product of
n linear
factors. We define the multiplicity of a root
r to be the number
of factors the polynomial has of the form x  r. 
Example:
Suppose that a polynomial passes through the point
(0,3) and has a the roots as follows:
Find the polynomial.
Solution:
The polynomial has the form
P(x) = a(x  i)^{2}(x + i)^{2}(x  (2 +
))^{3}(x
 (2  ))^{3}(x  4)^{2}
To find a we have that
3 = P(0) = a(1)(1)(1)(16)
Hence
a = 3/16
so that
P(x) = 3/16(x  i)^{2}(x + i)^{2}(x  (2 +
))^{3}(x  (2 
))^{3}(x  4)^{2}

The Rational Root Theorem
The Rational Root Theorem
Let
P(x) = a_{n}x^{n} +
a_{n1}x^{n1} + ... + a_{1}x +
a_{0}
and let p/q be a root of P(x) then
p is a positive or negative factor of a_{0} and
q is a
positive factor of a_{n}. 
Example: Factor
P(x) = 6x^{3} + x^{2} 
15x + 4
Solution:
The factors of 4 are 1,1, 2,2,4,4
The positive factors of 6 are 1,2,3,6
Hence the possibilities for rational roots are
1,1, 2,2,4,4,1/2,1/2,1/3,1/3,2/3,2/3,4/3,4/3.
Using the graph we see that the roots are near 1/3, 1/2, and 4/3. We try synthetic division until we find that 4/3 is a root of the equation
and that
6x^{2} + 9x  3
is the quotient of P(x) and x 
4/3.
We have
6x^{2} + 9x  3 = 3(2x^{2} + 3x  1)
Next we use the quadratic formula to find that the other two roots are
(3 +
)/4 and (3  )/4.
We can say that the factors are
x  3/4,
x  [(3 + )/4],
and x  [(3  /4]

Descartes' Rule of Signs
Decartes' Rule of Signs
Let P(x) be a polynomial with real coefficients:

The number of positive real roots either is equal to the number
of variations in the sign of P(x) or is less then that by an even integer.

The number of negative roots is either equal to the number of variations
in the sign of P(x) or is less than that by an even integer. Alternatively,
to compute the number of negative roots, you change the sign of all the odd
terms of P(x) and count the number of signs.

Example:
Consider the function
f(x) = x^{3} + 8x + 5
Since all the signs are positive, there are no sign changes, hence there
are no positive roots.
f(x) = x^{3}  8x + 5
We see one sign change hence there is exactly one negative real root. Therefore
there must be two imaginary roots. Using synthetic division, we have
that neither 5 nor 1 are roots, hence there are no rational roots. Since
there is only one negative root, it cannot be of the form
a +
b
We
can conclude that there is one irrational root that is not a square root,
and two other roots a + bi and a  bi.
Steps for finding roots:

Use Descartes' rule of signs to determine positive and negative
real roots.

Use the p/q theorem in coordination with Descartes' Rule of signs
to find a possible roots.

Plug in 1 and 1 to see if one of these two possibilities is a root.
If so go to step 5.

If not use synthetic division to test the other possibilities for
roots until you find the first root. If there are no rational roots, give up or use a calculator.

Find the quotient of dividing f(x) by x 
r.

If the quotient is a quadratic, use the quadratic formula to find
the last two roots. Otherwise go back to Step 2 applied to the quotient.

Reread the question and answer it.
Exercises
a) Find the rational zeros of each polynomial function
b) Factor each over the set of integers

x^{3} + 7x^{2} + 16x + 12

2x^{3}  9x^{2} + 14x  10

4x^{4}  28x^{3} + 67x^{2}  63x +
18

x^{5} + x^{4}  6x^{3} + 8x 
16
For a worked out example, click here