3.6: Zeros of Polynomial Functions
- Page ID
- 117120
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Evaluating a Polynomial Using the Remainder Theorem
In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by \(x–k\), the remainder may be found quickly by evaluating the polynomial function at \(k\), that is, \(f(k)\). Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend \(f(x)\) and a non-zero polynomial divisor \(d(x)\) where the degree of \(d(x)\) is less than or equal to the degree of \(f(x)\),there exist unique polynomials \(q(x)\) and \(r(x)\) such that
\[f(x)=d(x)q(x)+r(x) \nonumber\]
If the divisor, \(d(x)\), is \(x−k\), this takes the form
\[f(x)=(x−k)q(x)+r \nonumber\]
Since the divisor \(x−k\)
is linear, the remainder will be a constant, \(r\). And, if we evaluate this for \(x=k\), we have
\[\begin{align*} f(k)&=(k−k)q(k)+r \\[4pt] &=0{\cdot}q(k)+r \\[4pt] &=r \end{align*}\]
In other words, \(f(k)\) is the remainder obtained by dividing \(f(x)\)by \(x−k\).
If a polynomial \(f(x)\) is divided by \(x−k\),then the remainder is the value \(f(k)\).
Using the Factor Theorem to Solve a Polynomial Equation
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm.
\[f(x)=(x−k)q(x)+r\]
If \(k\) is a zero, then the remainder \(r\) is \(f(k)=0\) and \(f (x)=(x−k)q(x)+0\) or \(f(x)=(x−k)q(x)\).
Notice, written in this form, \(x−k\) is a factor of \(f(x)\). We can conclude if \(k\) is a zero of \(f(x)\), then \(x−k\) is a factor of \(f(x)\).
Similarly, if \(x−k\) is a factor of \(f(x)\), then the remainder of the Division Algorithm \(f(x)=(x−k)q(x)+r\) is \(0\). This tells us that \(k\) is a zero.
This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree \(n\) in the complex number system will have \(n\) zeros. We can use the Factor Theorem to completely factor a polynomial into the product of \(n\) factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.
According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\).
- Use synthetic division to divide the polynomial by \((x−k)\).
- Confirm that the remainder is \(0\).
- Write the polynomial as the product of \((x−k)\) and the quadratic quotient.
- If possible, factor the quadratic.
- Write the polynomial as the product of factors.
Using the Rational Zero Theorem to Find Rational Zeros
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial
Consider a quadratic function with two zeros, \(x=\frac{2}{5}\) and \(x=\frac{3}{4}\). By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor.
Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4.
We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.
The Rational Zero Theorem states that, if the polynomial \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) has integer coefficients, then every rational zero of \(f(x)\) has the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
- Determine all factors of the constant term and all factors of the leading coefficient.
- Determine all possible values of \(\dfrac{p}{q}\), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. Be sure to include both positive and negative candidates.
- Determine which possible zeros are actual zeros by evaluating each case of \(f(\frac{p}{q})\).
Using the Fundamental Theorem of Algebra
Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.
Suppose \(f\) is a polynomial function of degree four, and \(f (x)=0\). The Fundamental Theorem of Algebra states that there is at least one complex solution, call it \(c_1\). By the Factor Theorem, we can write \(f(x)\) as a product of \(x−c_1\) and a polynomial quotient. Since \(x−c_1\) is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it \(c_2\). So we can write the polynomial quotient as a product of \(x−c_2\) and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of \(f(x)\).
The Fundamental Theorem of Algebra states that, if \(f(x)\) is a polynomial of degree \(n > 0\), then \(f(x)\) has at least one complex zero.
We can use this theorem to argue that, if \(f(x)\) is a polynomial of degree \(n >0\), and a is a non-zero real number, then \(f(x)\) has exactly \(n\) linear factors
\[f(x)=a(x−c_1)(x−c_2)...(x−c_n)\]where \(c_1,c_2\),...,\(c_n\) are complex numbers. Therefore, \(f(x)\) has \(n\) roots if we allow for multiplicities.
Using the Linear Factorization Theorem to Find Polynomials with Given Zeros
A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have \(n\) zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into \(n\) factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form \((x−c)\), where c is a complex number.
Let \(f\) be a polynomial function with real coefficients, and suppose \(a +bi\), \(b≠0\), is a zero of \(f(x)\). Then, by the Factor Theorem, \(x−(a+bi)\) is a factor of \(f(x)\). For \(f\) to have real coefficients, \(x−(a−bi)\) must also be a factor of \(f(x)\). This is true because any factor other than \(x−(a−bi)\), when multiplied by \(x−(a+bi)\), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function \(f\) with real coefficients has a complex zero \(a +bi\), then the complex conjugate \(a−bi\) must also be a zero of \(f(x)\). This is called the Complex Conjugate Theorem.
According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form \((x−c)\), where \(c\) is a complex number.
If the polynomial function \(f\) has real coefficients and a complex zero in the form \(a+bi\), then the complex conjugate of the zero, \(a−bi\), is also a zero.
Given the zeros of a polynomial function \(f\) and a point \((c, f(c))\) on the graph of \(f\), use the Linear Factorization Theorem to find the polynomial function.
- Use the zeros to construct the linear factors of the polynomial.
- Multiply the linear factors to expand the polynomial.
- Substitute \((c,f(c))\) into the function to determine the leading coefficient.
- Simplify.
Using Descartes’ Rule of Signs
There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in \(f(x)\) and the number of positive real zeros. For example, the polynomial function below has one sign change.
This tells us that the function must have 1 positive real zero.
There is a similar relationship between the number of sign changes in \(f(−x)\) and the number of negative real zeros.
In this case, \(f(−x)\) has 3 sign changes. This tells us that \(f(x)\) could have 3 or 1 negative real zeros.
According to Descartes’ Rule of Signs, if we let \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) be a polynomial function with real coefficients:
- The number of positive real zeros is either equal to the number of sign changes of \(f(x)\) or is less than the number of sign changes by an even integer.
- The number of negative real zeros is either equal to the number of sign changes of \(f(−x)\) or is less than the number of sign changes by an even integer.