Download Unit 4 Lesson 1 Day 5

Review and Activating Strategy:
Complete the “Connect 4”
Worksheet
Essential Questions
How do we apply various theorems to find roots of polynomial functions
and to solve polynomial equations?
Standards
MM3A3: Students will solve a variety of equations and inequalities.
MM3A3a: Find real and complex roots of higher degree polynomial
equations using the factor theorem, remainder theorem, rational root
theorem, and fundamental theorem of algebra, incorporating complex and
radical conjugates.
MM3A3b: Solve polynomial, exponential, and logarithmic equations
analytically, graphically, and using appropriate technology.
MM3A3c: Solve polynomial, exponential, and logarithmic inequalities
analytically, graphically, and using appropriate technology. Represent
solution sets of inequalities using interval notation.
MM3A3d: Solve a variety of types of equations by appropriate means,
choosing among mental calculation, pencil and paper, or appropriate
technology.
Irrational and Imaginary Roots and Conjugate Pairs
Two other theorems can be helpful in finding the roots of a polynomial:
Irrational Roots Theorem: If P is a polynomial with rational coefficients and
a + b is a root of the polynomial where a and b are rational and is irrational,
then a - b is also a root.
Complex Conjugate Roots Theorem: If P is a polynomial in one variable with
real coefficients and a + bi is a root of P, then a – bi must also be a root of P.
In other words, these types of roots always come in pairs. (We could reason this from
the quadratic formula.)
Based on the Complex Conjugate Root Theorem, we can determine what combinations
of real and imaginary roots are possible for a polynomial if we know the degree. For
example, if we have a 5th degree polynomial, the following combinations would be
possible:
0 imaginary and 5 real
2 imaginary and 3 real
4 imaginary and 1 real
In every case, we have 5 roots; since the imaginary roots come in pairs, we couldn’t
have, for example, 1 imaginary and 4 real roots.
9.
For each polynomial below, state the number of real zeros (counting multiplicities)
and the possible combinations of real and imaginary numbers of roots.
a.
f(x) = x3 – 5x2 + 4x + 4
b. g(x) = x6 – 4x2 + 17 c. h(x) = x5 – 8x4 + 2x3 – 4x2 – 3
20
Let’s consider a scenario where some of the roots are imaginary. Suppose that
you were asked to find the roots of
f(x) = x4 – x3 + 3x2 – 4x – 4.
10
-10
-5
5
-10
Using the Rational Root Theorem, we can determine that there are only six
possible rational roots: . If we test all of these roots, we will find that none of
them leads to a remainder of 0, so there are no rational roots for this function.
We can also verify this by looking at the graph of this function; we can see on
the graph that none of the six possible rational solutions are x-intercepts of the
function.
-20
We know (from the Fundamental Theorem of Algebra) that the function must
have at least one root, so if there are no rational roots, the roots must be
irrational or imaginary. Without a little help at this point we are absolutely
stuck. None of the strategies we have discussed so far help us at this point.
Suppose that at this point we are given information that one of the roots of the
function is 2i. Because roots come in pairs, an additional root should be -2i. So, we can
take these values and use them for synthetic division.
Substitute 2i:
Substitute -2i :
2i 1
-1
3
-4
-4
-2i 1 -1+2i
-1 – 2i
-2i
1
1
Suppose that at this point we are given information that one of the roots of the
function is 2i. Because roots come in pairs, an additional root should be -2i. So, we can
take these values and use them for synthetic division.
Substitute 2i:
Substitute -2i :
2i 1
-1
3
-4
-4
-2i 1 -1+2i
-1 – 2i
-2i
2i
-4 – 2i 4 – 2i
4
-2i
+ 2i
2i
1
-1+2i -1 – 2i
-2i
0
1
-1
-1
0
The process is the same as the process we used testing possible rational roots,
though the multiplication step may be a little more complicated in the parts
that require distribution. We can see that these substitutions give remainders
of 0, and each substitution reduces the degree on the part of the polynomial
that needs to be divided into linear factors. After these two substitutions, we
can rewrite the polynomial as a product of two linear and one quadratic factor:
(x – 2i)(x + 2i)(x2 – x – 1)
We can use the quadratic formula to factor the quadratic part:
a = 1, b = -1, c = -1:
1  (1) 2  4(1)( 1) 1  1  4 1  5
x


2(1)
2
2
 1  5  1  5 
 x 
.
The complete factorization is f ( x)  ( x  2i )( x  2i ) x 



2
2



1 5 1 5
,
.
The roots are 2i, - 2i,
2
2
10. Use the given information to find the complete factorization and
name the roots for each polynomial below.
Show your work on a separate sheet of paper.
a. f(x) = x4 + 4x2 – 45, given that -3i is a root.
b. g(x) = x5 – x4 + 23x3 – 23x2 – 50x + 50
c. h(x) = x4 + 3x3 + 6x2 + 48x - 160
Remainder Theorem: Theorem that states that the remainder of a
polynomial f(x) divided by a divisor (x – t) is equal to f(t).
Factor Theorem: Theorem that states that a polynomial f(x) has a factor
(x – t) if and only if f(t) = 0.
Complex Conjugate Roots Theorem: Theorem that states that if P is a
polynomial in one variable with real coefficients and a + bi is a root of P,
then a – bi must also be a root of P.
Fundamental Theorem of Algebra: Theorem that states that every
non-zero single-variable polynomial with complex coefficients has
exactly as many complex roots as its degree, if each root is counted up
to its multiplicity.
11. To summarize, complete the table below to explain how the theorems discussed
in this task help us find the roots.
(If you need more space to write, you may re-create the table on a separate sheet.)
Theorem:
Remainder
Theorem
Factor Theorem
Complex Conjugate
Roots Theorem
Fundamental
Theorem of Algebra
In your own words:
How it helps us find roots of polynomials: