We saw in that topic what is called the factor theorem.
The Factor Theorem. x − r is a factor of a polynomial P(x) if and only if r is a root of P(x).
This means that if a polynomial can be factored, for example, as follows:
P(x) = (x − 1)(x + 2)(x + 3)
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:
(x + 2)(x − 1)(x − 5)
We will see how to prove the factor theorem below.
0. Because r being a root will mean that x − r is a factor
of p(x).
given that one
root is 3.
The three roots are: 2, −3, 3.
given that one root is −2.
Since −2 is a root, then (x + 2) is a factor. To find the other, quadratic factor, divide the polynomial by x + 2. Note that the root −2 goes in the box:
We have
x³ − 2x² − 5x + 6 |
= |
(x² − 4x + 3)(x + 2) |
|
|
= |
(x − 1)(x − 3)(x + 2) |
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 n > 2?
We must be given, or we must guess, a root r. We can then divide the polynomial by x − r, and hence produce a factor of the polynomial that will be one degree less. If we can discover a root of that polynomial, we can continue the process, reducing the degree each time, until we reach a quadratic, which we can always solve.
Here is a theorem that will help us guess a root.
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 factor of the constant term.
We will prove this below.
This Integer Root Theorem is an instance of the more general Rational Root Theorem:
If the rational number r/s is a root of a polynomial whose coefficients are integers, then the integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.
Example. What are the possible integer roots of x³ − 4x² + 2x + 4?
Answer. If there are integer roots, they will be factors of the constant term 4; namely: ±1, ±2, ±4.
Now, is 1 a root? To answer, we will divide the polynomial by x − 1, and hope for remainder 0.
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:
±1. Because those are the only factors of the constant term.
No, because neither 1 nor −1 will make that polynomial equal to 0. Synthetic division by both ±1 does not give remainder 0.
Conjugate pairs
If the irrational number a + is a root, then its conjugate a − is also a root. (See Skill in Algebra, Lesson 28.) And if the complex number a + bi is a root, then so is its conjugate, a − bi.
Example 6. A polynomial P(x) has the following roots:
−2, 1 + , 5i.
What is the smallest degree that P(x) could have?
Answer. 5. For, since 1 + is a root, then so is its conjugate, 1 − . And since 5i is a root, so is its conjugate, −5i.
P(x) has at least these 5 roots:
−2, 1 ± , ±5i.
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 x² − 4x + 1. .
b) 2 − 3i
Since 2 − 3i is a root, then so is 2 + 3i. Again, according to the theorem of the sum and product of the roots, they are the roots of x² − 4x + 13. See Topic 10, Example 7.
Problem 9. Let f(x) = x5 + x4 + x3 + x2 − 12x − 12. One root is and another is −2i.
a) Name the certain roots that f(x) has.
±, ±2i.
b) What are the possible integer roots?
The factors of the constant term −12:
±1, ±2, ±3, ±4, ±6, ±12.
c) Of those twelve possible roots, how many could f(x) actually have?
One. This is a polynomial of degree 5, which has 5 roots, and we already know 4 of them.
Problem 10. 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 x is a large negative number, the graph is below the x-axis. When x is a large positive number, it is above the x-axis. Therefore, the graph must cross the x-axis at least once. Now, can you draw the graph so that it crosses the x-axis exactly twice? No, you cannot. A polynomial of odd degree must have an odd number of real roots.
Proof of the factor theorem
x − r is a factor of a polynomial P(x)
if and only if
r is a root of P(x).
First, if (x − r) is a factor of P(x), then P(r) will have the factor (r − r), which is 0. This will make P(r) = 0. This means that r is a root.
Conversely, if r is a root of P(x), then P(r) = 0. But according to the remainder theorem, P(r) = 0 means that upon dividing P(x) by x − r, the remainder is 0. x − r, therefore, is a factor of P(x).
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 factor of the constant term.
Let the integer r be a root of this polynomial:
P(x) = ±xn + an−1xn−1 + an−2xn−2 + . . . + a2x2 + a1x + a0,
where the a's are integers. Then, since r is a root,
P(r) =
±rn +
an−1rn−1 + an−2rn−2 + . . . + a2r2 + a1r + a0 = 0.
Transpose the constant term a0, and factor r from the remaining terms:
r(±rn−1 + an−1rn−2 + . . . + a2r + a1) = −a0
Now the a's are all integers; therefore the expression in parentheses is an integer, which, for convenience, we will call −q:
r(−q) = −a0,
or,
rq = a0.
Thus, the constant term a0 can be factored as rq, if r and q are both integers. Under those conditions, then, r is a factor of the constant term.
This is what we wanted to prove.
Next Topic: Multiple roots
Table of Contents | Home
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2001-2007 Lawrence Spector
Questions or comments?
E-mail: themathpage@nyc.rr.com