Download Solving Poly. Eq.

Mathematical Investigations III
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Mathematical Investigations III - A View of the World
Solving Polynomial Equations
We observed in Worksheet 4 that the number of roots of a polynomial is at most the degree of
that polynomial (unless that polynomial is P(x) = 0). The number of roots can be much less than
the degree; consider P(x) = x1000, which has only zero as a root. In one sense, though, the number
of roots is equal to the degree.
Recall the definition of the multiplicity of a root from Worksheet 5.
1.
For each given polynomial, find each of its roots and the roots’ corresponding
multiplicities. Feel free to use your calculator.
2.
a.
x3 – 3x + 2
d.
x3 + 3x2 + 3x + 1
b.
x3 + 5x2 + 4x
e.
c.
(x – 1)2(x2 + 1)
x1000
What do you notice about the sum of the multiplicities of the roots of a polynomial?
Sometimes we refer to “summing the multiplicities of the roots” as “counting the roots according
to multiplicity.”
The observations above are consequences of the following very deep theorem.
Fundamental Theorem of Algebra
If P(x) is a polynomial with complex coefficients and positive degree n, then P(x) has
exactly n complex roots when counting according to multiplicity.
Poly 12.1
Rev. S11
Mathematical Investigations III
Name:
Poly 12.2
Rev. S11
Mathematical Investigations III
Name:
When trying to find all roots of a polynomial, it usually helps to find an easy root first. For
example, with the polynomial p(x) = x3 + x2 + 7x – 9, if we recognize that p(x) has a root at x = 1,
then we can factor out x – 1 to get p(x) = (x – 1)(x2 + 2x + 9). At this point, the quadratic formula
tells us that the other two roots of p(x) are  1  i  8 . This idea always works when we can find
a root. For if r is a root of P(x) so that P(r) = 0, then we can use the division algorithm to find
that P(x) = (x – r)Q(x) + R. Plug in x = r on both sides of the equation to learn that 0 = R. So
finding a root allows us to factor the polynomial and thus hopefully get to work with an easier
polynomial. What we need is a nice way to find the “easy roots.” Here’s one trick.
3.
Consider the polynomial equation x3 – x2 + 13x + 75 = 0.
a.
What is the product of the roots?
b.
list all integers that are factors of your answer to part a. Don’t forget negatives!
c.
If there is an integer root to the equation, why must it be among your list from part b.?
d.
Test all possibilities from part b. to see if any actually are roots.
e.
You should have found a root in part d. Divide the polynomial by (x – r) as in the
paragraph at the top of the page.
f.
Now find the other roots of the polynomial.
Poly 12.3
Rev. S11
Mathematical Investigations III
Name:
4.
Solve each of the polynomial equations below. Keep in mind that you might find more
than one root among your possible integer roots, or that one of them might be a “multiple root”
(that is, have multiplicity more than one).
a.
x4 – 6x3 + 3x2 + 24x – 28 = 0
b.
x4 – 3x3 – 12x – 16 = 0
c.
x4 – 13x2 + 20x – 4 = 0
Poly 12.4
Rev. S11
Mathematical Investigations III
Name:
Our trick breaks down if the leading coefficient of our polynomial is not one, because then we
have to divide the whole polynomial through by this coefficient before we can find the product of
the roots, and the product might not even be an integer. In this case, we expand the ideas above
using the:
Rational Roots Theorem
p
is a
q
rational number in lowest terms that is a root of P (x ) , then a n is divisible by q and a 0 is divisible
by p.
Let P(x)  an x n  an1x n1 
 a1x1  a0 be a polynomial with integer coefficients. If
What this is saying is that if there are any rational number solutions at all (and there may not be!)
the constant term of the polynomial must be a multiple of the numerator and the coefficient of the
highest degree term must be a multiple of the denominator. Otherwise, a fraction has no hope of
being a root of the polynomial.
5.
Solve 3x3 + 14x2 + 5x – 2 = 0.
a.
List all possible numerators of rational roots (there should be four):
b.
List all possible denominators of rational roots (there are four of these, too):
c.
List all possible rational roots of the polynomial (there should be eight):
d.
Test each possible rational number from part c. to see if it is a root.
e.
If you found a root (you should have!) divide the polynomial by the relevant factor.
(Hint: if, say, 3/7 is a root, you have the option of dividing by (x  73 ) or by (7x – 3).)
f.
Find the other roots of this polynomial equation.
Poly 12.5
Rev. S11
Mathematical Investigations III
Name:
6.
Solve 4x3 – 21x2+ 40x – 25 = 0. Yes, there are 18 possibilities for the rational roots.
Just as in the case of integer roots, there may be multiple different rational roots, or rational roots
of multiplicity higher than one. There are also plenty of polynomials that don’t have any rational
roots at all, and then you’re on your own—the rational roots theorem is more-or-less the best tool
available, so when it doesn’t work mathematicians are in uncharted waters.
More practice.
7.
Solve x3 + 22x + 348 = 0. (It may help to graph to find nice roots!)
8.
The equation x3 – 6x – 10 = 0 doesn’t have any rational roots. Use your calculator to find
all real roots, accurate to two decimal places. Can you think of more than one way to do this?
Poly 12.6
Rev. S11
Mathematical Investigations III
Name:
9.
10.
a.
Find the roots of P(x) = x4 + x3 – 10x2 – 4x + 24..
b.
Now solve the inequality x4 + x3 – 10x2 – 4x + 24 > 0.
a.
Solve x4 + 10x3 + 24x2 – 32x = 128.
b.
Solve for x: log(x)  10 log(x)  24 log(x)  32 log(x)  128.
4
3
Poly 12.7
2
Rev. S11
Mathematical Investigations III
Name:
Even More Bonus Material!
11. a.
Let f(x) = 2x3 – 11x2 + 15x – 1. Find f(2) and f(3). Explain why there must be a root
of f(x) between x = 2 and x = 3.
b. Try calculating f(2.5). Can you explain how you might use a divide-and-conquer
strategy to locate the root more accurately?
12. As mentioned, sometimes the rational root theorem works, sometimes it isn’t helpful. Find
all rational roots (including multiplicity!) of each polynomial below (don’t look for other roots):
a.
3x4 + 10x3 + 6x2 + 10x + 3
b.
3x3 + 5x2 + 4x – 3
c.
8x4 – 4x3 – 6x2 + 5x – 1.
Poly 12.8
Rev. S11
Mathematical Investigations III
Name:
For completeness, here’s a proof of the Rational Roots Theorem:
Rational Roots Theorem
p
is a
q
rational number in lowest terms that is a root of P (x ) , then a n is divisible by q and a 0 is divisible
by p.
Let P(x)  an x n  an1x n1 
Proof: The fact that
 a1x1  a0 be a polynomial with integer coefficients. If
p
is a root of P (x ) means that
q
n
 p
 p
 p
0  P   an    an1  
q
q
q
n1
1
 p
   a1    a0 ,
q
So
pn
p n1
p

a
   a1  a0 .
n1
n
n1
q
q
q
We can clear denominators by multiplying both sides of the equation by q n :
0  an p n  an1 p n1q    a1 pq n1  a0 q n .
0  an
By moving the a0 q n to the left-hand side and factoring on the right, we get
 a0 q n  p(an p n1  an1 p n2 q    a1q n1 ) ,
p
was
q
assumed to be in lowest terms. Therefore a 0 must, in fact, have a factor of p , and thus a 0 is
divisible by p .
Similarly, we could start with the third equation above and instead move an p n to the left-hand
side to argue that a n is divisible by q , but we will leave that problem as an exercise.
So  a0 q n has a factor of p . However, q cannot be divisible by p since the fraction
13. Complete the proof of the Rational Roots Theorem by demonstrating that a n must be
divisible by q .
In case you are curious, we cannot offer you a proof of the Fundamental Theoream of Algebra.
Though many proofs are known to mathematicians, there are none known that do not involve
calculus or another branch or higher mathematics. Now you have something to look forward to!
Poly 12.9
Rev. S11