Real Zeros of Polynomials
These are my notes on real zeros of polynomials.
In this section, we discuss the techniques that can be used to find the real
zeros of a polynomial function. Recall that if r is a real zero of a polynomial
function f then f(r)=0, r is an x-intercept of the graph of f, and r is a
solution of the equation f(x)=0.
For polynomial and rational functions, we have seen the importance of the real
zeros for graphing. In most cases, the real zeros of a polynomial function are
difficult to find using algebraic methods. No nice formulas like the quadratic
formula are available to help us find zeros for polynomials of degree 3 or
higher. Formulas do exist for solving any third or fourth degree polynomial
equation, but they are complicated to use. No general formulas exist for
polynomial equations of degree 5 or higher.
Remainder and Factor Theorems
When we divide one polynomial by another, we obtain a quotient polynomial and a
remainder, the remainder being either the zero polynomial or a polynomial whose
degree is less than the degree of the divisor. To check our work, we do:
(quotient)(divisor) + remainder = dividend
This checking routine is the basis for a famous theorem called the division
algorithm for polynomials.
Remainder Theorem
Let f be a polynomial function. If f(x) is divided by x-c, then the remainder is
f(c).
Example 1
Using the remainder theorem
Find the remainder if \(f(x)=x^3-4x^2-5\) when divided by x-3 and x+2
Solution:
We could use long division, but who wants to do that, it is easier to use the
remainder theorem.
\(f(3)=(3)^3-4(3)^2-5 = 27-36-5 = 14\)
The remainder is 14
To find the remainder when f(x) is divided by x+2 = x-(-2) = -2
\(f(-2) = (-2)^3-4(-2)^2-5 = -8-16-5 = -29\)
The remainder is -29
Factor Theorem
Let f be a polynomial function. Then x-c is a factor of f(x) if ands only if
f(c)=0. If f(c)=0, then x-c is a factor of f(x). If x-c is a factor of f(x),
then f(c)=0.
Suppose that f(c)=0. Then, we have \(f(x)=(x-c)q(x)\) for some polynomial q(x).
That is, x-c is a factor of f(x). Suppose that x-c is a factopr of f(x). Then
there is a polynomial function q such that \(f(x)=(x-c)q)x)\). Replacing x by c,
we find that \(f(c)=(c-c)q(c) = 0*q(c) = 0\). That completes the proof. One use
of the factor theorem is to determine whether a polynomial has a particular
factor.
Example 2
Use the factor theorem to determine whether a function has a factor
\(f(x)=2x^3-x^2+2x-3\)
For x-1 and x+3
Solution:
That factor theorem states that if f(c)=0 then x-c is a factor
Because x-1 is of the form x-c with c=1, we find the value of f(1). We choose to
use substitution.
\(f(1) = 2(1)^3-(1)^2+2(1)-3 = 2-1+2-3 = 0\)
By the factor theorem, x-1 is a factor of x.
To test the factor x+3, we first need to write it in the form of x-c. Since
x+3=x-(-3), we find the value of f(-3). Because f(-3) =-72 and not 0, we
conclude from the factor theorem that x-(-3) or x+3 is not a factor of f(x).
Number of Real Zeros
The next theorem concerns the number of real zeros that a polynomial function
may have. In counting the zeros of a polynomial, we count each zero as many
times as its multiplicity.
A polynomial function cannot have more real zeros than its degree.
The proof is based on the factor theorem. If r is a real zero of a polynomial
function f, then f(r)=0 and so, x-r is a factor of f(x). Each real zero
corresponds to a factor of degree 1. Because f cannot have more first degree
factors than its degree.
Descarte's rule of signs provides information about the number and location of
the real zeros of a polynomial function written in standard form. It requires
that we count the number of variations in the sign of the coefficients of f(x)
and f(-x). For example, the following polynomial function has two variations in
the signs of the coefficients.
\(f(x)=3x^7+4x^4+3x^2-2x-1 = -3x^7+0x^6+0x^5+4x^4+0x^3+3x^2-2x-1\)
Notice that we ignored the zero coefficients in \(0x^6,0x^5\) and \(0x^3\) in
counting the number of variations in the sign of f(x). Replacing x by -x, we
get:
\(f(-x)=-3(-x)^7+4(-x)^4+3(-x)^2-2(-x)-1 = 3x^7+4x^4+3x^2+2x-1\)
This has one variation in sign.
Descarte's Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of variations in
the sign of the nonzero coefficients of f(x) or else equals that number less an
even integer.
The number of negative real zeros of f either equals the number of variations in
the sign of the nonzero coefficients of f(-x) or else equals that number less an
even integer.
Example 3
Using the number of real zeros theorem and the rule of signs.
Discuss the real zeros of \(f(x)=3x^6-4x^4+3x^3+2x^2-x-3\)
Solution:
Because the polynomial is of degree 6, by the real zeros theorem there are at
most six real zeros. Since there are three variations in the sign of the nonzero
coefficients of f(x), by the rule of signs, we expect either three or one
positive real zero. To continue, look at f(-x).
\(f(-x)=3x^6-4x^4-3x^3+2x^2+x-3\)
There are three variations in sign, so we expect either three or one negative
real zeros. Equivalently, we know now that the graph of f has either three or one
positive x-intercepts and three or one negative x-intercepts.
Rational Zeros Theorem
The rational zeros theorem provides information about the rational zeros of a
polynomial with integer coefficients.
Let f be a polynomial function of degree 1 or higher of the form:
\(f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...a_{1}x+a_{0}\)
Each coefficient is an integer. If p/q is in lowest terms, it is a rational zero
of f, then p must be a factor of \(a_o\), and q must be a factor of \(a_n\).
Example 4
Listing potential rational zeros
\(f(x)=2x^3+11x^2-7x-6\)
Solution:
Because f has integer coefficients, we may use the rational zero theorem. First,
we list all the integers p that are factors of the constant term a0=-6 and all
the integers q that are factors of the leading coefficient ay=2.
\(p:....\pm1,\pm2,\pm3,\pm6\)
\(q:....\pm1,\pm2\)
Now, we form all possible ratios p/q
\(\frac{p}{q}:....\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)
If f has a rational zero, it will be found in this list, which contains 12
possibilities.
Be sure that you understand what the rational zeros theorem says. For a
polynomial with integer coefficients, if there is a rational zero, it is one of
those listed. It may be the case that the function does not have any rational
zeros.
Long division, synthetic division, or substitution can be used to test each
potential rational zero to determine whether it is indeed a zero. To make the
work easier, integers are usually tested first.
Example 5
Finding the rational zeros of a polynomial function
\(f(x)=2x^3+11x^2-7x-6\)
Write f in factored form
Solution:
We gather all the information that we can about the zeros.
1. There are at most 3 zeros
2. By the rule of signs, there is one positive real zero. Also, because
\(f(-x)=-2x^3+11x^2+7x-6\)
There are two negative zeros or no negative zeros.
3. Now we use the list of potential rational zeros obtained in the previous
example.
\(\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)
We choose to test the potential rational zero using substitution.
\(f(1)=2(1)^3+11(1)^2-7(1)-6 = 2+11-7-6 = 0\)
Since f(1)=0, 1 is a zero and x-1 is a factor of f. We can use long division or
synthetic division to factor f.
\(f(x)=2x^3+11x^2-7x-6 = (x-1)(2x^2+13x+6)\)
Now, any solution of the equation \(2x^2+13x+6=0\) will be a zero of f. Because
of this, we call the equation a depressed equation of f. Since the degree of the
depressed equation of f is less than that of the original polynomial, it is
easier to work with the depressed equation to find the zeros of f.
4. The depressed equation \(2x^2+13x+6=0\) is a quadratic equation with
discriminant \(b^2-4ac = 169-48 = 121 > 0\). The equation has two real
solutions, which can be found by factoring:
\(2x^2+13x+6 = (2x+1)(x+6) = 0\)
\(2x+1 = 0\) and \(x+6 = 0\)
So, \(x=-\frac{1}{2}\) or \(x = -6\)
The zeros of f are \(-6,-\frac{1}{2},1\)
We know that \(f(x)=2x^3+11x^2-7x-6\) can be written as
\(f(x)=(x-1)(2x^2+13x+6)\). Because \(2x^2+13x+6 = (2x+1)(x+6)\), we write f in
factored form as: \(f(x)=(x-1)(2x+1)(x+6)\)
Notice that the three zeros of f are found in this example are among those given
in the list of potential zeros.
Find the Real Zeros of a Polynomial Function
1. Use the degree of the polynomial to determine the maximum number of real
zeros.
2. Use the rule of signs to determine the possible number of positive zeros and
negative zeros.
3. If the polynomial has integer coefficients, use the rational zeros theorem to
identify those rational numbers that potentially could be zeros.
Use substitution, synthetic division, or long division to test each potential
rational zero. Each time that a zero(and thus a factor) is found, repeat step 3
on the depressed equation.
4. In attempting to find the zeros, remember to use the factoring techniques
that you already know.
Example 6
Finding the real zeros of a polynomial function
Find the real zeros of \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)
Write f in factored form.
Solution:
We gather all the information that we can about the zeros.
There are at most 5 real zeros because of \(x^5\).
By the rule of signs, there are 5, three, or one positive zero.
There are no negative zeros because \(f(x)=-x^5-5x^4-12x^3-24x^2-32x-16\) has no
variation in sign.
Because the leading coefficient \(a_{5}=1\) and there are no negative zeros, the
potential rational zeros are the integers 1,2,4,8, and 16, the positive factors
of the constant term, 16. We test the potential rational zero 1 first, using
synthetic division.
We get a remainder of f(1)=0, so 1 is a zero and x-1 is a factor of f. Using the
entries in the bottom row of the synthetic division, we can begin to factor f.
\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x^4-4x^3+8x^2-16x+16)\)
We now work with the first depressed equation:
\(q1(x)=x^4-4x^3+8x^2-16x+16=0\)
The potential rational zeros of q1 are still 1,2,4,8, and 16. We test 1 first
since it may be a repeated zero of f.
Since the remainder is 5, 1 is not a repeated zero. We try 2 next:
The remainder is f(2)=0 so 2 is a zero and x-2 is a factor of f. Again using the
bottom row, we find:
\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)(x^3-2x^2+4x-8)\)
The remaining zeros satisfy the new depressed equation:
\(q2(x)=x^3-2x^2+4x-8=0\)
Notice that q2(x) can be factored using grouping.
Since \(x^2+4=0\) has no real solutions, the real zeros of f are 1 and 2, the
latter being a zero of multiplicity 2. The factored form of f is:
\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)^2(x^2+4)\)
Example 7
Solving a polynomial equation
Find the real solutions of the equation:
\(x^5-5x^4+12x^3-24x^2+32x-16=0\)
Solution:
The real solutions of this equation are the real zeros of the polynomial
function \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)
Using the result from previous example, the real zeros of f are 1 and 2. These
are the real solutions of the equation:
\(x^5-5x^4+12x^3-24x^2+32x-16=0\)
In example 6, the quadratic factor \(x^2+4\) that appears in the factored form
of f is called irreducible, because the polynomial \(x^2+4\) cannot be factored
over the real numbers. In general, we say that a quadratic factor \(ax^2+bx+c\)
is irreducible if it cannot be factored over the real numbers, that is, if it is
prime over the real numbers.
Refer to examples 5 and 6. The polynomial function of example 5 has three real
zeros, and its factored form contains three linear factors. The polynomial
function of example 6 has two distinct real zeros, and its factored form
contains two distinct linear factors and one irreducible quadratic factor.
Every polynomial function with real coefficients can be uniquely factored into a
product of linear factors and/or irreducible quadratic factors.
We shall prove this result and we shall draw several additional conclusions
about the zeros of a polynomial function. One conclusion is worth noting. If a
polynomial with real coefficients is of odd degree, it must contain at least one
linear factor. This means it must have at least one real zero.
A polynomial function with real coefficients of odd degree has at least one real
zero.
Theorem for Bounds on Zeros
The search for the real zeros of a polynomial function can be reduced somewhat
if bounds on the zeros are found. A number M is a bound on the zeros of a
polynomial if every zero lies between -M and M. That is, M is a bound on the
zeros of a polynomial if \(-M \leq\text{ any real zero of f }\leq M\)
Example 8
Using the theorem for finding bounds on zeros
Find a bound on the real zeros of each polynomial
1. \(f(x)=x^5+3x^3-9x^2+5\)
2. \(g(x)=4x^5-2x^3+2x^2+1\)
Solution:
1. The leading coefficient of f is 1.
\(f(x)=x^5+3x^3-9x^2+5 = 0,3,-9,0 = 5\)
The smaller of the two numbers, 10, is the bound. Every real zero of f lies
between -10 and 10.
2. First, we write g so that it is the product of a constant times a polynomial
whose leading coefficient is 1.
\(g(x)=4x^5-2x^3+2x^2+1 = 4(x^5-\frac{1}{2}x^3+\frac{1}{2}x^2+\frac{1}{4}\)
Next, we evaluate the two expressions in (4) with
\(a_4=0,a_3=-\frac{1}{2},a_2=\frac{1}{2},a_1=0\text{ and }a_0=\frac{1}{4}\)
The smaller of the two numbers, \(\frac{5}{4}\) is the bound. Every real zero
of g lies between \(-\frac{5}{4}\text{ and }\frac{5}{4}\)
Intermediate Value Theorem
It is based on the fact that the graph of a polynomial function is continuous,
that is, it contains no holes or gaps. Let f denote a polynomial function. If
a<b and if f(a) and f(b) are of opposite sign, there is at least one real zero
of f between a and b.
Example 9
Using the intermediate value theorem to locate real zeros
Show that \(f(x)=x^5-x^3-1\) has a real zero between 1 and 2.
Solution:
We evaluate f at 1 and 2
\(f(1)=-1\)
\(f(2)=23\)
Because f(1)<0 and f(2)>0, it follows from the intermediate value theorem that f
has a zero between 1 and 2.
Let us look at the polynomial f of example 9 more closely. Based on the rule of
signs, f has exactly one positive real zero. Based on the rational zeros
theorem, 1 is the only potential positive rational zero. Since f(1) does not
equal 0, we conclude that the zero between 1 and 2 is irrational. We can use the
intermediate value theorem to approximate it.
1. Find two consecutive integers a and a+1 such that f has a zero between them.
2. Divide the interval into 10 equal subintervals.
3. Evaluate f at each endpoint of the subintervals until the intermediate value
theorem applies. This endpoint then contains a zero.
4. Repeat the process until the desired accuracy is achieved.
Example 10
Approximating the real zeros of a polynomial function
Find the positive zero of:
\(f(x)=x^5-x^3-1\) to two decimal places.
Solution:
From example 9, we know that the positive zero is between 1 and 2. We divide the
interval [1,2] into 10 equal subintervals.
[1,1.1][1.1,1.2][1.2,[1.3][1.3,1.4][1.4,1.5][1.5,1.6][1.6,1.7][1.7,1.8][1.9,1.9][1.9,2].
Now we find the value of f at each endpoint until the intermediate value theorem
applies.
\(f(x)=x^5-x^3-1\)
f(1.0)=-1 f(1.2)=-0.23968 f(1.1)=-0.72049 f(1.3)=0.51593
We can stop here and conclude that the zero is between 1.2 and 1.3. now we
divide the interval[1.2,1.3] into 10 equal subintervals and proceed to evaluate
f at each endpoint. We conclude that the zero lies between 1.23 and 1.24 and
correct to two decimal places, the zero is 1.23.