#### Transcript 6.5 Theorems About Roots of Poly Equations

```5.5 Theorems About Roots of
Poly Equations
The Rational Root Theorem
If p/q is in simplest form and is a rational
root of a polynomial equation, then p must
be a factor of a0 and q must be a factor of
an.
This says that any roots must be a factor of
the coefficient of the last term over a factor
of the coefficient of the first term.
Example:
Find all possible roots of x3-4x2-2x+8 = 0.
We need factors of 8 over factors of 1.
Factors of 8: plus or minus 1,2,4,8
Factors of 1: plus or minus 1
The possible factors of the poly are plus or
minus 1/1,2/1,4/1,8/1.
We can then check these to find actual roots
using synthetic division.
Conjugate Root Theorem
Let a and b be rational numbers and let √b
be an irrational number. If a+ √b is a root
of a poly then the conjugate a- √b is also a
root.
Example: If 1+ √3 and - √11 are roots give
me two more roots?
Imaginary Root Theorem
If the imaginary number a+bi is a root of a
poly, then the conjugate a-bi is also a root.
Example: Find two more roots if 3i and -2+i
are both roots.
Desartes’ Rule of Signs
Let P(x) be a poly with real coefficients
written in standard form.
• The number of positive real roots of P(x) =
0 is either equal to the number of sign
changes between consecutive coefficients
of P(x) or is less than that by an even
number.
• The number of negative real roots is either
equal to the number of sign changes
between consecutive coefficients of P(-x)
or is less than that by an even number.
• WHAT?
Example:
Tell about the nature of the roots of
x3 – x2 + 1 = 0
There are two positive sign changes, so we
know that there are either 2 or 0 positive
roots.
There is only one negative sign change so
we know that we only have one negative
root.
We can use these theorems to rewrite the
original poly.
Example: Find a 3rd degree poly with roots --1,2-i.
We know that (x+1)(x-(2-i))(x-(2+i)) are the
factors.
Now foil and distribute to find the poly.
(x+1)(x-(2-i))(x-(2+i))=
foil
(x+1)(x2-x(2-i)-x(2+i)+(2-i)(2+i))=
simplify
(x+1)(x2-2x+xi-2x-xi+4-2i+2i-i2)=
simplify
(x+1)(x2-4x+5)=
distribute
x(x2-4x+5)+1(x2-4x+5)=
simplify
x3-4x2+5x+x2-4x+5= x3-3x2+x+5=0
HW pg 316 9 , 12 , 18, 23, 24, 31, 37
Lets do 37 now
```