13 ## ROOTS OF POLYNOMIALS## OF DEGREE GREATER THAN 2The fundamental theorem of algebra Proof of the integer root theorem IN THIS TOPIC we will see how to find the roots of a polynomial of degree greater than 2. This will depend on the previous topic: Synthetic division. We saw in that topic what is called the factor theorem.
The Factor Theorem. This means that if a polynomial can be factored, for example, as follows:
then the theorem tells us that the roots are 1, −2, and −3. Conversely, if we know that roots of a polynomial are −2, 1, and 5, then the polynomial has the following factors: ( We could then multiply out and know the polynomial that has those three roots. We will see below how to prove the factor theorem . a) Use the Factor Theorem to prove: ( To see the answer, pass your mouse over the colored area.
−1 is a root of b) Use synthetic division to find the other factor.
Therefore, Following this same procedure, we could prove: ( and completely generally: ( The Fundamental Theorem of Algebra The following is called the Fundamental Theorem of Algebra:
This apparently simple statement allows us to conclude:
If the leading coefficient of
Hence a polynomial of the third degree, for example, will have three roots. And if they are all real, then its graph will look something like this: For, the three roots are the three
As for a polynomial of the fourth degree, it will have four roots. And if they are all real, then its graph will look something like this:
Here, the graph on the far left is Example 1. Write the polynomial with integer coefficients that has the following roots: −1, ¾.
The factors are (4 The polynomial is 4 Problem 2. Determine the polynomial whose roots are −1, 1, 2, and sketch its graph. The factors are (
Here is the graph: The Problem 3. Determine the polynomial with integer coefficients whose roots are −½, −2, −2, and sketch the graph. The factors are (2 2 Here is the graph: −2 is a double root. The graph does not cross the
Question. If
0. Since
Problem 4. Is
Use synthetic division to divide the polynomial by The remainder is 0. 2 is a root of the polynomial. Example 2. Find the three roots of
given that one root is 3.
We have
The three roots are: 2, −3, 3. Again, since Problem 5. Sketch the graph of this polynomial,
given that one root is −2. Since −2 is a root, then ( We have
The three roots are: 1, 3, −2. Here is the graph: A strategy for finding roots What, then, is a strategy for finding the roots of a polynomial of degree We must be given, or we must guess, a root Here is a theorem that will help us guess a root.
The integer root theorem. We will prove this below. This Integer Root Theorem is an instance of the more general Rational Root Theorem:
Example 3. What are the possible integer roots of
Now, is 1 a root? To answer, we will divide the polynomial by
The remainder is not 0. 1 is not a root. Let's try −1:
The remainder again is not 0. Let's try 2:
Yes! 2 is a root. We have
We can now find the roots of the quadratic by completing the square. As we found in Topic 11:
Therefore, the three roots are: 1 + , 1 − , 2. Problem 6. a) What are the possible integer roots of this polynomial?
±1. They are the only factors of the constant term. b) Does that polynomial have integer roots? No, because neither 1 nor −1 will make that polynomial equal to 0. Synthetic division by both ±1 does not give remainder 0. Problem 7. Factor this polynomial into a product of linear factors.
We must find the roots. The possible integer roots are ±1, ±2, ±3, ±6. Synthetic division reveals that −1 is a root. Therefore,
Conjugate pairs If the irrational number
Example 4. A polynomial −2, 1 + , 5 What is the smallest degree that
−2, 1 ± , ±5 Problem 8. Construct a polynomial that has the following root: a) 2 +
Since 2 + is a root, then so is 2 − . Therefore, according to the theorem of the sum and product of the roots (Topic 10), they are the roots of b) 2 − 3
Since 2 − 3
Problem 9. Construct a polynomial whose roots are 1 and 5
Since 5
Next, since 1 is a root, then (
(
Problem 10. Let If
One. This is a polynomial of the 5th degree, and has 5 roots. Two are and −. And two are 2 Problem 11. Is it possible for a polynomial of the 5th degree to have 2 real roots and 3 imaginary roots? No, it is not. Since imaginary roots always come in pairs, then if there are any imaginary roots, there will always be an even number of them. Consider the graph of a 5th degree polynomial with positive leading term. When Proof of the factor theorem
First, if ( Conversely, if This is what we wanted to prove. Proof of the integer root theorem If an integer is a root of a polynomial whose coefficients are integers and whose leading coefficient is ±1, then that integer is a Let the integer
where the
Transpose the constant term
Now the
or,
Thus, the constant term This is what we wanted to prove. Please make a donation to keep TheMathPage online. Copyright © 2015 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |