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ALGEBRA II
LESSON 7-5
Pg. 371-377
ROOTS AND ZEROS
Today’s Key Concepts: What is the difference between the terms “root” and “zero”? The root is a solution of
an equation (a value of x that makes the equation true) which is also the location the graph of the function
crosses the x-axis known as an x-intercept, where the value of f(x) which you can think of as “y” = 0.
TERMS: 1. root: the solution of a quadratic equation (found in the form x = 2, x = – 3) 2 and – 3 are roots
2. zero: the x-intercept of the graph of a quadratic equation; the point(s) for which f(x) = 0
3. real number: all the numbers used in everyday life, both rational and irrational together
4. imaginary number: the principle square root of a negative number (also known as “i”)
5. complex number: any number that can be written in the form a + bi, where a and b are real
numbers and “i” is the imaginary unit
6. complex conjugate: (a + bi) and (a – bi)
A. We all should know that if we factor a trinomial like x2 – 5x –24 = 0 the “factors” that result are:
(x – 8)(x + 3) = 0
B. We also should know that if we “solve” this equation, we find the value of x when the factors are set equal
to zero. (the “zero of the function”) x – 8 = 0
x+3=0
x=8
x=–3
C. Therefore we have two solution or “roots” of this equation and both are real numbers: 8 and – 3
Key Concept: This Lesson will help us determine the number of roots an equation has and the type or
roots they are, with a few interesting methods of determining these values. Don’t get lost with remembering
all the concepts, just try to use as many as possible to become familiar with the processes.
THE FUNDAMENTAL THEOREM OF ALGEBRA
** Remember this: The degree of a polynomial is exactly the same as the number of roots or solutions it
contains. Example: f(x) = x4 – 16x3 + 2x2 – x + 3 has a degree of 4 and therefore has 4 roots
- the roots are complex roots meaning either real or imaginary or both.
DESCARTES RULE OF SIGNS
Rene Descartes, the French math guru, discovered in the 1600s an interesting trait that polynomials present
when looking at the sequence of positive and negative sign changes within the polynomial.
- First arrange the polynomial in descending order of the power of the variable
-- the number of positive real zeros is determined by counting the number of sign changes of the
coefficient that occur within the polynomial (or it may be less than this by 2 or 4 or 6…)
-- the number of negative real zeros is determined the same way by determining the number of
sign changes of the sister function f(-x) ** Simply reverse the sign of only the odd degree terms
and count the changes again! The number is the number of changes or less by 2 or 4 or 6…!
** Remember there can be “imaginary roots” too so don’t forget all the options below our number!
Example of “Rule of Signs”
f(x) = 8x3 – 6x2 – 23x + 6
yes
no yes
f(-x) = – 8x3 – 6x2 + 23x + 6
no yes
no
** 2 or 0 positive roots
** 1 Negative root (we must have one since we cannot have –1)
Since this a 3rd degree polynomial, it has 3 roots—2 or 0 positive roots and 1 negative root.
Our possible combinations of roots include: 2 positive and 1 negative = 3
0 positive and 1 negative and 2 imaginary =3
**NOTE: The number of imaginary zero comes from conjugate pairs and therefore must be an even
number.
Synthetic Substitution to find the Zeros of a Function.
This is the hard part and involves a lot of trial and error, but knowing the value of “Rule of Signs” and
Synthetic Substitution, we should be able to figure these out.
Let’s do Problem #33 to demonstrate this solution process.
r(x) = x4 – 6x3 + 12x2 + 6x –13
yes
yes no yes
r(-x) = x4 + 6x3 + 12x2 - 6x –13
no
no
yes no
Possibilities: 4 total zeros: 3 or 1 positive, 1 negative
So: 3 Positive with 1 Negative OR 1 Positive, 1 Negative, and 2 Imaginary
Now we pick random values of roots for synthetic substitution.
x :
-1
0
1
2
3
1 -6
1 -7
1
1
1
-5
-4
-3
12 6 - 13
19 -13
0
-13
7 13
0
4 14 15
3 15 32
x : 1 - 7 19 -13
1 : 1 -6 13 0
Solutions: +1, -1, 3 + 2i, and 3 – 2i
So we know the –1 is our negative root
Also + 1 is a positive root, suspect 2 imaginary roots
Let’s factor out the two knowns to see what remains:
If x +1 is factored: x3 - 7x2 + 19x –13 remains
Now factor x –1 from x3 - 7x2 + 19x –13 syntetically
So x –1 factored out leaves x2 – 6x + 13 which has imaginary roots
a=1
____________
______
b = -6
x = 6 + 36 – 4(1)(13) = 6 + 36 – 52 = 3 + 2i
c = 13
2
2
REVERSE THE PROCESS…KNOWING ZEROS, DETERMINE THE POLYNOMIAL FUNCTION
Steps to follow:
1. Convert the zeros into factors (Given 4 and 4 – i)
First understand imaginary factors have COMPLEX CONJUGATES because of (+) in the formula
So with “4 – i” comes “4 + i”
x=4
x = (4 – i)
x = (4 + i)
x–4=0
x - (4 – i) = 0
x - (4 + i) = 0
2. Multiply the factors together piece by piece:
[x - (4 – i)][x - (4 + i)] = [x – 4 + i][x – 4 – i] = [(x-4)2 – i2] = x2 – 8x + 16 – (-1) = x2 – 8x + 17
Take the results times the last zero:
(x2 – 8x + 17)(x – 4) = x3 – 8x2 + 17x – 4x2 + 32x – 68 = x3 – 12x2 + 49x - 68