3.05 Zeros of polynomial functions
Concept summary
The zeros of a function are the input values which make the function equal to zero. This means that zeros are the solutions to the equation f\left(x\right) = 0.
The multiplicity of a zero is the number of times that its corresponding factor appears in the function.
Describes the trend of a function or graph at its left and right ends; specifically the y-value that each end obtains or approaches
A point where a line or graph intersects the x-axis. The value of y is 0 at this point.
A point where a line or graph intersects the y-axis. The value of x is 0 at this point.
Worked examples
Example 1
Consider the function f(x)=2x^4+5x^3-11x^2-20x+12.
Determine all the zeros of f(x) and their multiplicities.
Approach
To find the zeros of the function f(x), we set f(x)=0 and solve this equation for x. Each value of x is a zero of the function.
To determine the multiplicities of each zero, we determine the number of times that the corresponding factor of each zero occurs in the function.
Solution
Note that if a number k is a zero of f(x), then f(k)=0. It follows that x-k is a factor of f(x).
The factors of the constant term which is 12 are \pm1,\pm2,\pm3,\pm4,\pm6,\pm12 and the factors of the leading coefficient which is 2 are \pm1,\pm2. We divide the factors of 12 and the factors of 2, so, the possible zeros of f(x) are \pm \dfrac{1}{2},\pm1,\pm \dfrac{3}{2},\pm2,\pm3,\pm4,\pm6 and \pm12.
Observe that
- f\left(\dfrac{1}{2}\right)=2\left(\dfrac{1}{2}\right)^4+5\left(\dfrac{1}{2}\right)^3-11\left(\dfrac{1}{2}\right)^2-20\left(\dfrac{1}{2}\right)+12=0
- f\left(2\right)=2\left(2\right)^4+5\left(2\right)^3-11\left(2\right)^2-20\left(2\right)+12=0
- f\left(-2\right)=2\left(-2\right)^4+5\left(-2\right)^3-11\left(-2\right)^2-20\left(-2\right)+12=0
- f\left(-3\right)=2\left(-3\right)^4+5\left(-3\right)^3-11\left(-3\right)^2-20\left(-3\right)+12=0
Therefore the zeros of f(x)=2x^4+5x^3-11x^2-20x+12 are x=\dfrac{1}{2},x=2,x=-2 and \\x=-3.
Moreover, the corresponding factors of each zero are the following:
- x=\dfrac{1}{2} corresponds to the factor 2x-1
- x=2 corresponds to the factor x-2
- x=-2 corresponds to the factor x+2
- x=-3 corresponds to the factor x+3
This implies that f(x)=(2x-1)(x-2)(x+2)(x+3).
Since the multiplicity of each zero is the number of times its corresponding factor appears in f(x), we have that:
- x=\dfrac{1}{2} is a zero with multiplicity 1
- x=2 is a zero with multiplicity 1
- x=-2 is a zero with multiplicity 1
- x=-3 is a zero with multiplicity 1
Reflection
Alternatively, we can find the zeros of f(x) by expressing the left-hand side of the equation 2x^4+5x^3-11x^2-20x+12=0 by its factors. That is,
2x^4+5x^3-11x^2-20x+12=(2x-1)(x-2)(x+2)(x+3)=0
Then we get the equations
2x-1=0,x-2=0,x+2=0,x+3=0
Therefore, the zeros and their mutiplicities are the following:
- x=\dfrac{1}{2} is a zero with multiplicity 1
- x=2 is a zero with multiplicity 1
- x=-2 is a zero with multiplicity 1
- x=-3 is a zero with multiplicity 1
Determine the end behavior of f\left(x\right) as x\to\infty.
Approach
We determine first the leading coefficient of f(x)=2x^4+5x^3-11x^2-20x+12 and its degree either odd or even since the end behavior of a polynomial function is determined by its degree and leading coefficient.
Solution
The function f(x)=2x^4+5x^3-11x^2-20x+12 has a positive leading coefficient which is 2.
Note that if the polynomial function has a positive leading coefficient, the end behavior is as follows:
| Even degree | Odd degree |
|---|---|
| \text{As }x\to -\infty, f(x)\to \infty. | \text{As }x\to -\infty, f(x)\to -\infty. |
| \text{As }x\to \infty, f(x)\to \infty. | \text{As }x\to \infty, f(x)\to \infty. |
The degree of f(x) is 4 which is even.
Therefore, as x\to \infty, f(x)\to \infty.
Reflection
Another way of determining the end behavior of f(x)=2x^4+5x^3-11x^2-20x+12 as x\to \infty is through looking its graph.
Clearly, as x\to \infty, f(x)\to \infty.
Determine the end behavior of f\left(x\right) as x\to -\infty.
Approach
We determine first the leading coefficient of f(x)=2x^4+5x^3-11x^2-20x+12 and its degree either odd or even since the end behavior of a polynomial function is determined by its degree and leading coefficient.
Solution
The function f(x)=2x^4+5x^3-11x^2-20x+12 has a positive leading coefficient which is 2.
Note that if the polynomial function has a positive leading coefficient, the end behavior is as follows:
| Even degree | Odd degree |
|---|---|
| \text{As }x\to -\infty, f(x)\to \infty. | \text{As }x\to -\infty, f(x)\to -\infty. |
| \text{As }x\to \infty, f(x)\to \infty. | \text{As }x\to \infty, f(x)\to \infty. |
The degree of f(x) is 4 which is even.
Therefore, as x\to -\infty, f(x)\to \infty.
Reflection
Another way of determining the end behavior of f(x)=2x^4+5x^3-11x^2-20x+12 as \\x\to -\infty is through looking its graph.
Clearly, as x\to -\infty, f(x)\to \infty.
Example 2
Consider the function f(x)=x^5+7x^4+17x^3+47x^2+72x-144.
Determine all the zeros of f(x) and their multiplicities.
Approach
To find the zeros of the function f(x), we set f(x)=0 and solve this equation for x. Each value of x is a zero of the function.
To determine the multiplicities of each zero, we determine the number of times that the corresponding factor of each zero occurs in the function.
Solution
The factors of the constant term which is -144 are \pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm9,\pm12,\pm16,\pm18,\pm24,\pm36,\pm48,\pm72,\pm144 and the factors of the leading coefficient which is 1 are \pm1. We divide the factors of -144 and the factors of 1, so, the possible rational zeros of f(x) are \pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm9,\pm12,\pm16,\pm18,\pm24,\pm36,\pm48,\pm72,\pm144.
Observe that
- f(-4)=(-4)^5+7(-4)^4+17(-4)^3+47(-4)^2+72(-4)-144=0
- f(1)=(1)^5+7(1)^4+17(1)^3+47(1)^2+72(1)-144=0
This means that x=-4 and x=1 are zeros of f(x). By substitution, we can verify that x=-4 and x=1 are the only rational zeros of f(x). We need to find the other zeros of the function which can be irrational or complex.
Note that if a number k is a zero of f(x), then f(k)=0. It follows that x-k is a factor of f(x). So, we have x+4 and x-1 as factors of f(x)=x^5+7x^4+17x^3+47x^2+72x-144. Observe that f(x)=x^5+7x^4+17x^3+47x^2+72x-144=(x+4)(x-1)(x^3+4x^2+9x+36).
We apply factoring by grouping for x^3+4x^2+9x+36:
| \displaystyle x^3+4x^2+9x+36 | \displaystyle = | \displaystyle \left(x^3+4x^2\right)+(9x+36) | Split based on common factors |
| \displaystyle = | \displaystyle x^2\left(x + 4\right) + 9\left(x + 4\right) | Factor out the GCF: x^2 and 9 | |
| \displaystyle = | \displaystyle \left(x+4\right)\left(x^2+9\right) | Factor out the common binomial factor |
Note that a^2+b^2=(a+bi)(a-bi). Then x^2+9=(x)^2+(3)^2=(x+3i)(x-3i).
Therefore, f(x)=(x-1)(x+4)^2(x+3i)(x-3i).
Consequently,
- x=1 is a zero with multiplicity 1
- x=-4 is a zero with multiplicity 2
- x=-3i is a zero with multiplicity 1
- x=3i is a zero with multiplicity 1
Reflection
We can check whether the zeros of f(x) are correct by multiplying their corresponding factors.
| \displaystyle f(x) | \displaystyle = | \displaystyle (x-1)(x+4)^2(x+3i)(x-3i) | Function expressed by its factors |
| \displaystyle = | \displaystyle (x-1)(x+4)^2(x^2+9) | Apply (a-bi)(a+bi)=a^2+b^2 | |
| \displaystyle = | \displaystyle (x-1)(x^2+8a+16)\left(x^2+9\right) | Apply (a+b)^2=a^2+2ab+b^2 | |
| \displaystyle = | \displaystyle x^5+7x^4+17x^3+47x^2+72x-144 | Simplify |
This suggests that the zeros and their multiplicities we obtained are correct.
Determine the end behavior of f\left(x\right) as x\to\infty.
Approach
We determine first the leading coefficient of f(x)=x^5+7x^4+17x^3+47x^2+72x-144 and its degree either odd or even since the end behavior of a polynomial function is determined by its degree and leading coefficient.
Solution
The function f(x)=x^5+7x^4+17x^3+47x^2+72x-144 has a positive leading coefficient which is 1.
Note that if the polynomial function has a positive leading coefficient, the end behavior is as follows:
| Even degree | Odd degree |
|---|---|
| \text{As }x\to -\infty, f(x)\to \infty. | \text{As }x\to -\infty, f(x)\to -\infty. |
| \text{As }x\to \infty, f(x)\to \infty. | \text{As }x\to \infty, f(x)\to \infty. |
The degree of f(x) is 5 which is odd.
Therefore, as x\to \infty, f(x)\to \infty.
Reflection
Another way of determining the end behavior of f(x)=x^5+7x^4+17x^3+47x^2+72x-144 as x\to \infty is through looking its graph.
Clearly, as x\to \infty, f(x)\to \infty.
Determine the end behavior of f\left(x\right) as x\to -\infty.
Approach
We determine first the leading coefficient of f(x)=x^5+7x^4+17x^3+47x^2+72x-144 and its degree either odd or even since the end behavior of a polynomial function is determined by its degree and leading coefficient.
Solution
The function f(x)=x^5+7x^4+17x^3+47x^2+72x-144 has a positive leading coefficient which is 1.
Note that if the polynomial function has a positive leading coefficient, the end behavior is as follows:
| Even degree | Odd degree |
|---|---|
| \text{As }x\to -\infty, f(x)\to \infty. | \text{As }x\to -\infty, f(x)\to -\infty. |
| \text{As }x\to \infty, f(x)\to \infty. | \text{As }x\to \infty, f(x)\to \infty. |
The degree of f(x) is 5 which is odd.
Therefore, as x\to -\infty, f(x)\to -\infty.
Reflection
Another way of determining the end behavior of f(x)=x^5+7x^4+17x^3+47x^2+72x-144 as x\to -\infty is through looking its graph.
Clearly, as x\to -\infty, f(x)\to -\infty.
Example 3
Sketch a graph of the function f(x)=-3(x+3)(x-1)^3(x+2)(2x+1)^2, labeling the intercepts.
Approach
To label the y-intercept in the graph, we first find the value of f(x) at x=0.
To label the x-intercept in the graph, we first find the value of x such that f(x)=0. The x-values of the x-intercepts are the zeros of the function.
Solution
Let x=0. Substituting x=0 into the function, we have f(0)=-3(0+3)(0-1)^3(0+2)\left(2(0)+1\right)^2=18 The coordinates of the y-intercept is (0,18).
Let f(x)=0. Then we have -3(x+3)(x-1)^3(x+2)(2x+1)^2=0 This implies that x+3=0,x-1=0,x+2=0 and 2x+1=0 which gives x=-3,x=1,\\x=-2, and x=-\dfrac{1}{2}, respectively. The coordinates of the x-intercepts are (-3,0),(-2,0), \left(-\dfrac{1}{2},0\right) and (1,0).
The graph of the function is shown below.
Example 4
A polynomial function f(x) has the following characteristics:
Degree of 6
Coefficient of x^6 is 2
Zeros are x=3 with multiplicity 2, x=-5 with multiplicity 2, x=\sqrt{2} with multiplicity 1 and \\x=-\sqrt{2} with multiplicity 1
- As x\to\infty, f(x)\to\infty.
- As x\to-\infty, f(x)\to\infty.
Draw a graph of the function f(x).
Approach
To graph the polynomial function f(x), we first find this function in factored form based on the given characteristics.
Solution
The zeros of the function and their multiplicities are x=3 with multiplicity 2, x=-5 with multiplicity 2, x=\sqrt{2} with multiplicity 1 and x=-\sqrt{2} with multiplicity 1. So, the corresponding factors are x-3 which appears twice, x+5 which appears twice, and x-\sqrt{2} and x+\sqrt{2} which both appear once. From these, we get the expression (x-3)^2(x+5)^2\left(x-\sqrt{2}\right)\left(x+\sqrt{3}\right)
Clearly, the degree of the product of the factors is 6. Since the coefficient of x^6 is 2, f(x)=2(x-3)^2(x+5)^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right) is possibly the polynomial function.
We will first check the end behavior of f(x)=2(x-3)^2(x+5)^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right). The leading coefficient of f(x) is positive which is 2.
Note that if the polynomial function has a positive leading coefficient, the end behavior is as follows:
| Even degree | Odd degree |
|---|---|
| \text{As }x\to -\infty, f(x)\to \infty. | \text{As }x\to -\infty, f(x)\to -\infty. |
| \text{As }x\to \infty, f(x)\to \infty. | \text{As }x\to \infty, f(x)\to \infty. |
The degree of the function is 6 which is even.
Therefore, as x\to\infty and as x\to-\infty, we have f(x)\to\infty.
We have verified that the polynomial function is f(x)=2(x-3)^2(x+5)^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)
A graph of the function is shown below.
Outcomes
MA.912.AR.1.1
Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.
MA.912.AR.1.8
Rewrite a polynomial expression as a product of polynomials over the real or complex number system.
MA.912.AR.6.5
Sketch a rough graph of a polynomial function of degree 3 or higher using zeros, multiplicity and knowledge of end behavior.