If $P(x)$ is a polynomial such that $P(\alpha)=0$, then we call $\alpha$ a zero of the polynomial $P(x)$. If $\alpha$ is a zero of $P(x)$, then $(x-\alpha)\mid P(x)$. Show We can then write $P(x)=(x-\alpha)R(x)$ with $\deg(P)>\deg(R)$. Repeating this process, we can continue factoring out $(x-\alpha)$ until we reach the point: $P(x)=(x-\alpha)^n Q(x)$ where $Q(\alpha)\neq 0$. This value of $n$ is what we call the multiplicity of the zero $\alpha$. In simpler terms, once you write your polynomial in factored form and combining like factors, the exponents are the multiplicities for the corresponding roots. Here, we have $P(x)=-2x^3-x^2+x=-2(x-0)^1(x-(-1))^1(x-\frac{1}{2})^1$ according to your work. We see then that $0$ is a zero of multiplicity one, $-1$ is a zero of multiplicity one, and $\frac{1}{2}$ is a zero of multiplicity one. Learning Objectives We have already explored the local behavior (the location of \(x\)- and \(y\)-intercepts) for quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general. Identify Zeros and Their Multiplicities from a GraphGraphs behave differently at various \(x\)-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off. Suppose, for example, we graph the function \[f(x)=(x+3)(x−2)^2(x+1)^3 \nonumber\] Notice in the figure to the right illustrates that the behavior of this function at each of the \(x\)-intercepts is different.The \(x\)-intercept −3 is the solution of equation \((x+3)=0\). The graph passes directly through the \(x\)-intercept at \(x=−3\). The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function. The \(x\)-intercept 2 is the repeated solution of equation \((x−2)^2=0\). The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. The factor is repeated, that is, \((x−2)^2=(x−2)(x−2)\), so the solution, \(x=2\), appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, \(x=2\), has multiplicity 2 because the factor \((x−2)\) occurs twice. The \(x\)-intercept −1 is the repeated solution of factor \((x+1)^3=0\). The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function \(f(x)=x^3\). We call this a triple zero, or a zero with multiplicity 3. Graphical Behavior of Polynomials at \(x\)-intercepts If a polynomial contains a factor of the form \((x−h)^p\), the behavior near the \(x\)-intercept is determined by the power \(p\). We say that \(x=h\) is a zero of multiplicity \(p\).
For zeros with even multiplicities, the graphs touch or are tangent to the \(x\)-axis. For zeros with odd multiplicities, the graphs cross or intersect the \(x\)-axis. See the figure below for examples of graphs of polynomial functions with a zero of multiplicity 1, 2, and 3. The graphs clearly show that the higher the multiplicity, the flatter the graph is at the zero. For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the \(x\)-axis. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the \(x\)-axis. How to: Given a graph of a polynomial function, identify the zeros and their mulitplicities
Example \(\PageIndex{1}\): Find Zeros and Their Multiplicities From a Graph Use the graph of the function of degree 6 in the figure below to identify the zeros of the function and their possible multiplicities.
Try It \(\PageIndex{2}\) Use the graph of the function in the figure below to identify the zeros of the function and their possible multiplicities.
Graph of a polynomial function. AnswerThe zero at -5 is odd. Since the curve is somewhat flat at -5, the zero likely has a multiplicity of 3 rather than 1. Find Zeros and their Multiplicities from a Polynomial EquationRecall that if \(f\) is a polynomial function, the values of \(x\) for which \(f(x)=0\) are called zeros of \(f\). If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros. How to: Given an equation of a polynomial function, identify the zeros and their multiplicities
Example \(\PageIndex{3}\): Find zeros and their multiplicity from a factored polynomial Find the zeros and their multiplicity for the following polynomial functions. a) \(f(x) = x(x+1)^2(x+2)^3\) b) \(f(x)=x^2(x^2-3x)(x^2+4)(x^2-x-6)(x^2-7) \) Solution. a) This polynomial is already in factored form. All factors are linear factors.
b) This polynomial is partly factored. All the zeros can be found by setting each factor to zero and solving
Now we need to count the number of occurrences of each zero thereby determining the multiplicity of each real number zero.
Try It \(\PageIndex{4}\) Find the zeros and their multiplicity for the polynomial \(f(x)=x^4-x^3−x^2+x\). AnswerZeros \(-1\) and \(0\) have odd multiplicity of \(1\). Zero \(1\) has even multiplicity of \(2\) Graph Polynomial FunctionsWe can use what we have learned about multiplicities, end behavior, and intercepts to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions. How to: Given a polynomial function, sketch the graph
Example \(\PageIndex{5}\): Sketch the Graph of a Polynomial Function Sketch a graph of \(f(x)=−2(x+3)^2(x−5)\).
Example \(\PageIndex{6}\) Sketch a graph of \(f(x)=x^2(x^2−1)(x^2−2)\). State the end behaviour, the \(y\)-intercept, and \(x\)-intercepts and their multiplicity. Solution Step 1. The leading term is the product of the high order terms of each factor: \( (x^2)(x^2)(x^2) = x^6\). Step 2. The \(y\)-intercept can be found by evaluating \(f(0)\). So \(f(0)=0^2(0^2-1)(0^2-2)=(0)(-1)(-2)=0 \). The \(x\)-intercepts can be found by solving \(f(x)=0\). Set each factor equal to zero. \[\begin{align*} x^2&=0 & & & (x^2−1)&=0 & & & (x^2−2)&=0 \\ x^2&=0 & &\text{ or } & x^2&=1 & &\text{ or } & x^2&=2 \\ x&=0, \:x=0 &&& x&={\pm}1 &&& x&={\pm}\sqrt{2} \end{align*}\] . This gives us five \(x\)-intercepts: \( (0,0)\), \((1,0)\), \((−1,0)\), \((\sqrt{2},0)\), and \((−\sqrt{2},0)\). Step 3. The polynomial is an even function because \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. The graph appears below. Graph of \(f(x)=x^2(x^2−1)(x^2−2)\) .Example \(\PageIndex{7}\) Sketch a graph of the polynomial function \(f(x)=x^4−4x^2−45\). State the end behaviour, the \(y\)-intercept, and \(x\)-intercepts and their multiplicity. Solution Step 1. The leading term is \(x^4\). Step 2. The \(y\)-intercept occurs when the input is zero. \[ \begin{align*} f(0) &=(0)^4−4(0)^2−45 =−45 \end{align*}\] The \(y\)-intercept is \((0,−45)\). The \(x\)-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial. \[\begin{align*} f(x)&=x^4−4x^2−45 \\ &=(x^2−9)(x^2+5) \\ &=(x−3)(x+3)(x^2+5) \[0=(x−3)(x+3)(x^2+5) \nonumber \] \( \begin{array}{ccccc} The \(x\)-intercepts \((3,0)\) and \((–3,0)\) all have odd multiplicity of 1, so the graph will cross the \(x\)-axis at those intercepts. Step 3. The polynomial is an even function because \(f(-x)=f(x)\), so the graph is symmetric about the y-axis. The graph appears below. The imaginary zeros are not \(x\)-intercepts, but the graph below shows they do contribute to "wiggles" (truning points) in the graph of the function. Graph of \(f(x)=x^4-4x^2-45\).Try It \(\PageIndex{8}\) Sketch a graph of \(f(x)=\dfrac{1}{6}(x−1)^3(x+3)(x+2)\). AnswerFigure \(\PageIndex{8}\): Graph of \(f(x)=\frac{1}{6}(x−1)^3(x+3)(x+2)\)Turning PointsDefinition: Turning Points A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree \(n\) will have, at most, \(n\) \(x\)-intercepts and \(n−1\) turning points. The degree of a polynomial function helps us to determine the number of \(x\)-intercepts and the number of turning points. A polynomial function of \(n\)th degree is the product of \(n\) factors, so it will have at most \(n\) roots or zeros. When the zeros are real numbers, they appear on the graph as \(x\)-intercepts. The graph of the polynomial function of degree \(n\) can have at most \(n–1\) turning points. This means the graph has at most one fewer turning points than the degree of the polynomial or one fewer than the number of factors. Figure \(\PageIndex{9}\): Graph of \(f(x)=x^4-x^3-4x^2+4x\),a 4th degree polynomial function with 3 turning points The maximum number of turning points of a polynomial function is always one less than the degree of the function. Example \(\PageIndex{9}\): Find the Maximum Number of Turning Points of a Polynomial Function Find the maximum number of turning points of each polynomial function.
Solution a. \(f(x)=−x^3+4x^5−3x^2+1\) First, rewrite the polynomial function in descending order: \(f(x)=4x^5−x^3−3x^2+1\) Identify the degree of the polynomial function. This polynomial function is of degree 5. The maximum number of turning points is \(5−1=4\). b. \(f(x)=−(x−1)^2(1+2x^2)\) First, identify the leading term of the polynomial function if the function were expanded: multiply the leading terms in each factor together. \( \begin{array}{rl} Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is \(4−1=3\). Example \(\PageIndex{10}\): Find the Maximum Number of Intercepts and Turning Points of a Polynomial Without graphing the function, determine the maximum number of \(x\)-intercepts and turning points for \(f(x)=−3x^{10}+4x^7−x^4+2x^3\). Solution The polynomial has a degree of \(n\)=10, so there are at most 10 \(x\)-intercepts and at most 9 turning points. Try It \(\PageIndex{11}\) Without graphing the function, determine the maximum number of \(x\)-intercepts and turning points for \(f(x)=108−13x^9−8x^4+14x^{12}+2x^3\) AnswerThere are at most 12 \(x\)-intercepts and at most 11 turning points. Example \(\PageIndex{12}\): Drawing Conclusions about a Polynomial Function from the Factors Given the function \(f(x)=−4x(x+3)(x−4)\), determine the \(y\)-intercept and the number, location and multiplicity of \(x\)-intercepts, and the maximum number of turning points. Solution The \(y\)-intercept is found by evaluating \(f(0)\). \[\begin{align*} f(0)&=−4(0)(0+3)(0−4)=0 \end{align*}\] The \(y\)-intercept is \((0,0)\). The \(x\)-intercepts are found by determining the zeros of the function. \( \begin{array}{ccc} The three \(x\)-intercepts \((0,0)\),\((–3,0)\), and \((4,0)\) all have odd multiplicity of 1. The degree is 3 so the graph has at most 2 turning points. Try It \(\PageIndex{13}\) Given the function \(f(x)=0.2(x−2)(x+1)(x−5)\), determine the local behavior. AnswerThe function is a 3rd degree polynomial with three \(x\)-intercepts \((2,0)\), \((−1,0)\), and \((5,0)\) all have multiplicity of 1, the \(y\)-intercept is \((0,2)\), and the graph has at most 2 turning points. Example \(\PageIndex{14}\): Drawing Conclusions about a Polynomial Function from the Graph
Try It \(\PageIndex{15}\) What can we conclude about the polynomial represented by the graph shown below based on its intercepts and turning points? A polynomial function AnswerThe end behavior indicates an odd-degree polynomial function (ends in opposite direction), with a negative leading coefficient (falls right). There are 3 \(x\)-intercepts each with odd multiplicity, and 2 turning points, so the degree is odd and at least 3. Write a Formula for a Polynomial given its GraphNow that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an \(x\)-intercept where each factor is equal to zero, we can form a function that will pass through a set of \(x\)-intercepts by introducing a corresponding set of factors. Note: Factored Form of Polynomials If a polynomial of lowest degree \(p\) has horizontal intercepts at \(x=x_1,x_2,…,x_n\), then the polynomial can be written in the factored form: \(f(x)=a(x−x_1)^{p_1}(x−x_2)^{p_2}⋯(x−x_n)^{p_n}\) where the powers \(p_i\) on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor \(a\) can be determined given a value of the function other than the \(x\)-intercept. How to: Given a graph of a polynomial function, write a formula for the function.
Example \(\PageIndex{16}\): Writing a Formula for a Polynomial Function from the Graph Construct the factored form of a possible equation for each graph given below.
Try It \(\PageIndex{17}\): Construct a formula for a polynomial given a graph Given the graph shown in Figure \(\PageIndex{21}\), write a formula for the function shown. Figure \(\PageIndex{18}\).Answer\(f(x)=−\frac{1}{8}(x−2)^3(x+1)^2(x−4)\) Try It \(\PageIndex{18}\): Construct a formula for a polynomial given a description Write a formula for a polynomial of degree 5, with zeros of multiplicity 2 at \(x\) = 3 and \(x\) = 1, a zero of multiplicity 1 at \(x\) = -3, and vertical intercept at (0, 9) Answer\(f(x) = \dfrac{1}{3} (x - 1)^2 (x - 3)^2 (x + 3)\) Key Concepts
Glossarymultiplicity Contributors
How do you find the zeros and multiplicity of a zero?If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. The sum of the multiplicities is the degree n.
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