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3. Polynomial Expressions, Equations, & Functions
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Algebra 2

3.05 Zeros and factors

Lesson
Practice

Zeros and factors

The zeros of a function are the input values which make the function equal to zero. This means a is a zero of f\left(x\right) if f\left(a\right) = 0. We also refer to these solutions as roots of the equation f\left(x\right)=0.

Fundamental theorem of algebra

A polynomial of degree n where n\geq 1 has a total of n roots in the set of complex numbers.

The fundamental theorem of algebra says the number of roots, including complex and repeated solutions, of any polynomial is equal to the degree of the polynomial. Remember that complex roots include the real and imaginary roots. The real zeros of a function will be the x-intercepts of its graph.

We can use the factor theorem to make connections between the roots and factors of a polynomial function.

Factor theorem

If x=a is a root of the polynomial equation f\left(x\right) = 0, then \left(x-a\right) is a factor of f\left(x\right).

The multiplicity of a zero is the number of times that its corresponding factor appears in the function. The multiplicities of the zeros in the function will sum to the degree of the polynomial by the fundamental theorem of algebra. Zeros with different multiplicities look different graphically.

x
y
Zeros of multiplicity 1
x
y
Zeros of multiplicity 2
x
y
Zeros of multiplicity 3
x
y
Graph of y=(x+3)(x+1)^{2}(x-2)^{3}

A root of multiplicity 1 crosses through the x-axis with no point of inflection (turn). A root with an odd multiplicity greater than 1 crosses through the x-axis with a point of inflection (turn). Roots with even multiplicity are tangent to the axis which means they touch the x-axis, then change direction and do not cross the x-axis.

Exploration

Match each graph to its equation.

  • y=\frac{1}{2}\left(x+1\right)\left(x-2\right)\left(x+3\right)
  • y=\frac{1}{4}\left(x+3\right)\left(x+1\right)\left(x-3\right)^{2}
  • y=-\left(x-2\right)\left(x+1\right)^{2}
  • y=\left(x-2\right)^{2}\left(x+1\right)^{2}
-4
-3
-2
-1
1
2
3
4
x
-8
-6
-4
-2
2
4
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8
y
\text{}
-4
-3
-2
-1
1
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x
-8
-6
-4
-2
2
4
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8
y
\text{}
-4
-3
-2
-1
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x
-4
-3
-2
-1
1
2
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y
\text{}
-4
-3
-2
-1
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x
-6
-5
-4
-3
-2
-1
1
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y
\text{}

  1. Explain how you found the correct equation for each graph.

  2. What are the similarities and differences between each of the graphs?

  3. What are the similarities and differences between each of the equations?

  4. If the roots are known, what other information would we need to know to find the equation for a specific function?

There are many polynomial functions that have the same roots. To find the equation of a specific function, we need to know:

  • the roots

  • the multiplicity of each root

  • the degree of the function

  • another point on the graph (to find the leading coefficient)

The leading coefficient, the real roots, the imaginary roots, and their multiplicities are what determines the exact equation of a function.

When the coefficients of a polynomial meet certain criteria, complex roots and irrational roots will come in conjugate pairs.

Complex conjugate roots theorem

If P\left(x\right) is a polynomial with real coefficients and a+bi is a root of P\left(x\right), then its complex conjugate a-bi is also a root.

Irrational conjugate roots theorem

If P\left(x\right) is a polynomial with rational coefficients and a+\sqrt{b} is an irrational root of P\left(x\right), then its conjugate a-\sqrt{b} is also a root.

Examples

Example 1

Given the function f(x)=-2(x-4)(3x+2)^{2}

a

State the degree of the polynomial.

Worked Solution
Create a strategy

The degree of a polynomial is the highest exponent on the independent variable when written in standard form.

Apply the idea

Since the polynomial is given in factored form, it will need to be rewritten in standard form to determine its degree.

\displaystyle f(x)\displaystyle =\displaystyle (-2x+8)(3x+2)^{2}Distribute -2 to first binomial
\displaystyle =\displaystyle (-2x+8)(9x^{2}+12x+4)Square (3x+2) using the sum of squares identity
\displaystyle =\displaystyle -18x^{3}+48x^{2}+88x+32Multiply polynomial expressions

Since the highest exponent in f(x) is 3, the degree of the function is 3.

Reflect and check

Since the degree is only related to the independent variable, we can ignore any coefficients or constant terms when calculating the degree from a factored polynomial. \begin{aligned}-2(x-4)(3x+2)^{2}&\to(x)(x)^{2}&=x^3\end{aligned}

The degree of the polynomial provides information about the number of x-intercepts, end behavior, range, and turning points of the graph. Prior to graphing, we would expect f(x) to have at most 3 \, x-intercepts, up to 2 inflection points, a range of (-\infty, \infty), and no absolute extrema.

b

Determine the zeros of the polynomial and their multiplicities

Worked Solution
Create a strategy

To determine the zeros of a function, each unique factor is set equal to zero and solved. The multiplicity of a zero is the number of times that the zero's factor appears in a function.

Apply the idea

To solve for the zeros, each unique factor of f(x) will be set equal to zero and solved for x.

\displaystyle x-4\displaystyle =\displaystyle 0Set factor equal to zero
\displaystyle x\displaystyle =\displaystyle 4Solve for x
\displaystyle 3x+2\displaystyle =\displaystyle 0Set factor equal to zero
\displaystyle 3x\displaystyle =\displaystyle -2
\displaystyle x\displaystyle =\displaystyle -\dfrac{2}{3}Solve for x

In the factored form of f(x), we see that (x-4) appears once and (3x+2) appears twice. So, f(x) has a zero of 4 with a multiplicity of 1, and a zero of -\dfrac{2}{3} with a multiplicity of 2.

Reflect and check

The multiplicities of the zeros give additional information about the graph of f(x). A zero with an even multiplicity will have the graph change directions without passing through the x-axis, while a zero with an odd multiplicity will pass through the x-axis.

The function f(x) will have x-intercepts at \left(-\dfrac{2}{3},0\right) and (4,0). At \left(-\dfrac{2}{3},0\right), the graph touches the x-axis, then changes directions. At (4,0), the graph crosses through the x-axis.

Also, since the fundamental theorem of algebra says that the number of real or complex solutions is equal to the degree of the polynomial, and our number of real zeros including repeating solutions equals the degree, we know there are no complex (non-real) zeros to f(x).

Example 2

Consider the graph of a cubic function shown below. Determine the equation of the function.

-5
-4
-3
-2
-1
1
2
3
x
-40
-30
-20
-10
10
20
30
40
f\left(x\right)
Worked Solution
Create a strategy

To determine the equation, we need to determine the zeros and their multiplicities first. We can use that information with the factor theorem to write the factors of the function.

The zeros and factors are not enough to determine the exact equation of this specific function though, so we will use the given point \left(-2,-27\right) to find the leading coefficient.

Apply the idea

From the graph, we can determine the zeros by finding the x-intercepts. These are at x = -3 and x=-\dfrac{1}{2}. The graph crosses through the x-axis at x=-3, so it must have a multiplicity of 1. The graph is tangent at x=-\dfrac{1}{2}, and the degree of the function is 3, so this zero has a multiplicity of 2.

Using the factor theorem with the zeros and their multiplicities, we have the expression \left(x+3\right)\left(x+\dfrac{1}{2}\right)^2 Alternatively, we can express a factor with a zero of -\dfrac{1}{2} as \left(2x+1\right): \left(x+3\right)\left(2x+1\right)^{2} Next, we need to use the given point to find the leading coefficient.

\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x+3\right)\left(2x+1\right)^2Equation of the function
\displaystyle -27\displaystyle =\displaystyle a\left(-2+3\right)\left(2\cdot-2+1\right)^2Substitute values from the given point
\displaystyle -27\displaystyle =\displaystyle a\left(1\right)\left(-3\right)^2Evaluate the multiplication and addition
\displaystyle -27\displaystyle =\displaystyle 9aEvaluate the multiplication
\displaystyle -3\displaystyle =\displaystyle aDivide both sides by 9

Therefore, f\left(x\right)=-3\left(x+3\right)\left(2x+1\right)^2 is the cubic function.

Reflect and check

To verify this answer, we can use the equation to find another point and check that it lies on the graph shown. The y-intercept is an easy point to verify, so we will substitute x=0 into the equation.

\displaystyle f\left(x\right)\displaystyle =\displaystyle -3\left(x+3\right)\left(2x+1\right)^2Equation of the function
\displaystyle f\left(0\right)\displaystyle =\displaystyle -3\left(0+3\right)\left(2\cdot 0+1\right)^2Substitute x=0
\displaystyle =\displaystyle -3\left(3\right)\left(1\right)^2Evaluate the multiplication and addition
\displaystyle =\displaystyle -9Evaluate the multiplication

Looking at the graph, we see that the y-intercept is at \left(0,-9\right), so the equation we found is correct.

Example 3

A polynomial function f\left(x\right) has the following characteristics:

  • Degree of 3

  • Zeros include x=3 and x=2i

  • Real coefficients

  • Has a y-intercept at \left(0,-12\right)

Determine the equation of the function.

Worked Solution
Create a strategy

The zeros of the function that were given are x=3 and x=2i. The latter is an imaginary root, so we need to determine if we can use the complex conjugates theorem by looking at the information given about the coefficients. The coefficients must be real in order to use the theorem.

Apply the idea

Since x=2i is a zero and the polynomial has real coefficients, then x=-2i must also be a zero by the complex conjugates theorem. From these, we get the expression \left(x-3\right)\left(x-2i\right)\left(x+2i\right)

The degree of the product of the factors is 3. Next, we need to use the y-intercept to find the leading coefficient.

\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x-3\right)\left(x-2i\right)\left(x+2i\right)Equation of the function
\displaystyle -12\displaystyle =\displaystyle a\left(0-3\right)\left(0-2i\right)\left(0+2i\right)Substitute values from the y-intercept
\displaystyle -12\displaystyle =\displaystyle a\left(-3\right)\left(-2i\right)\left(2i\right)Evaluate the addition
\displaystyle -12\displaystyle =\displaystyle -12aEvaluate the multiplication
\displaystyle 1\displaystyle =\displaystyle aDivide both sides by -12

Therefore,f\left(x\right)=\left(x-3\right)\left(x-2i\right)\left(x+2i\right) is the polynomial function, which can also be written as f(x)=(x-3)(x^{2}+4), or f(x)=x^{3}-3x^{2}+4x-12.

Reflect and check

If the information did not tell us that the coefficients were real, then we could not have used the complex conjugates theorem. This theorem can only be used if the coefficients are real since imaginary solutions must come in pairs and the imaginary components would disappear when factors are expanded.

Example 4

For each given function, determine the zeros using appropriate methods and state the multiplicity of each zero.

a

g(x)=3x^{3}-15x^{2}+24x-12

Worked Solution
Create a strategy

The factor theorem connects the roots of a polynomial to its factors, so factoring the polynomial (if possible) will allow us to determine the zeros. How often each factor appears in the function will also tell us the multiplicity of each zero.

Apply the idea

Replace g(x) with 0 and factor the polynomial.

\displaystyle 0\displaystyle =\displaystyle 3x^{3}-15x^{2}+24x-12Set g(x)=0
\displaystyle 0\displaystyle =\displaystyle 3(x^{3}-5x^{2}+8x-4)Take out a greatest common factor

From here, we need to find a way to break up one of the terms so we can factor by grouping. For this one, we can rewrite 8x into 4x+4x which will allow us to find another factor.

\displaystyle 0\displaystyle =\displaystyle 3(x^{3}-5x^{2}+4x+4x-4)Rewrite since 8x=4x+4x
\displaystyle 0\displaystyle =\displaystyle 3\left[x\left(x^2-5x+4\right)+4\left(x-1\right)\right]Factor by grouping
\displaystyle 0\displaystyle =\displaystyle 3\left[x\left(x-4\right)\left(x-1\right)+4\left(x-1\right)\right]Factor the quadratic
\displaystyle 0\displaystyle =\displaystyle 3\left[\left(x-1\right)\left(x\left(x-4\right)+4\right)\right]Factor GCF of \left(x-1\right)
\displaystyle 0\displaystyle =\displaystyle 3(x-1)(x^{2}-4x+4)Simplify the factors
\displaystyle 0\displaystyle =\displaystyle 3(x-1)(x-2)^{2}Factor completely

Once completely factored, set each factor equal to zero and solve.

0=x-2 \qquad 0=x-1

2=x \qquad 1=x

We see that the factor (x-2) appears twice and the factor (x-1) appears once. The zeros are 2 with a multiplicity of 2, and 1 with a multiplicity of 1.

Reflect and check

Since the degree of the polynomial is 3 and the multiplicities of the zeros also add to 3, we know that there will not be any complex (non-real) solutions to g(x).

b

f(x)=-3x^{3}+12x^{2}-3x

Worked Solution
Create a strategy

After setting the function equal to zero, we will determine if the polynomial can be factored, taking out a greatest common factor if possible. The quadratic formula will be used where needed if a prime quadratic appears after factoring. The number of each zero will determine the multiplicity.

Apply the idea

Replace f(x) with 0 and determine if the polynomial can be factored.

\displaystyle 0\displaystyle =\displaystyle -3x^{3}+12x^{2}-3xSet f(x)=0
\displaystyle 0\displaystyle =\displaystyle -3x(x^{2}-4x+1)Take out a greatest common factor

Since x^{2}-4x+1 can not be factored, we will use the quadratic formula, x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}, using values of a=1, b=-4, and c=1.

\displaystyle x\displaystyle =\displaystyle \dfrac{4 \pm \sqrt{(-4)^{2}-4(1)(1)}}{2(1)}Replace values of a,\, b,\, c in quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{4 \pm \sqrt{12}}{2}Evaluate b^{2}-4ac and 2a
\displaystyle x\displaystyle =\displaystyle \dfrac{4 + 2\sqrt{3}}{2},\,\dfrac{4 - 2\sqrt{3}}{2} Simplify radicals and write as separate solutions
\displaystyle x\displaystyle =\displaystyle 2+\sqrt{3},\,2-\sqrt{3}Simplify

Since -3x is a factor of f(x), the remaining zero will be the solution to -3x=0, or x=0. Each zero appears once, so the zeros 0 ,\, 2-\sqrt{3},\, and 2+\sqrt{3} all have multiplicities of 1.

Reflect and check

Since f(x) had a greatest common factor with a variable, we knew that there was at least one real solution. The value of the discriminant, b^{2}-4ac,\, lets us know whether the remaining solutions will be both real, both complex, or one of each.

The value of the discriminant of x^{2}-4x+1 is positive, which means that both remaining solutions of f(x) are real. Since the degree of the polynomial is 3, that means that there would be no complex zeros to the polynomial f(x).

Example 5

Show that the fundamental theorem of algebra is true for quadratic functions.

Worked Solution
Create a strategy

We can use the quadratic formula and the discriminant to explain the different types of solutions there are to a quadratic equation, then show that each case satisfies the fundamental theorem of algebra.

Apply the idea

All quadratic functions have a degree of 2, so they all have 2 zeros according to the fundamental theorem of algebra. The zeros of any quadratic function can be found by the quadratic formula x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} According to the discriminant, there are only 3 types of solutions that a quadratic equation can have.

  • When b^2-4ac>0, there are 2 real solutions.

  • When b^2-4ac=0, there is 1 real solution.

  • When b^2-4ac<0, there are 2 imaginary solutions.

x
y

The first case says there are 2 real solutions.

Graphically, we can see that each zero has a multiplicity of 1 because the graph crosses through the x-axis.

Therefore, the quadratic has 2 solutions in the set of complex numbers.

x
y

The second case says there is 1 real solution.

Graphically, we can see that this zero has a multiplicity of 2 because it is tangent to the x-axis.

This means the quadratic is in the form {y=k\left(x-\alpha\right)^2} which can also be written as y=k\left(x-\alpha\right)\left(x-\alpha\right), where k and \alpha are real numbers.

Therefore, the quadratic has 2 solutions in the set of complex numbers.

x
y

The final case says there are 2 imaginary solutions.

Graphically, this means there are no x-intercepts. However, if we extend to a complex plane, rotate the graph 90 \degree and then reflect the graph, it will intersect the complex x-plane twice.

Recall the set of complex numbers contains both real and imaginary solutions. Therefore, the quadratic has 2 solutions in the set of complex numbers.

This shows that the fundamental theorem of algebra is true for quadratic functions.

Idea summary

The fundamental theorem of algebra says that a polynomial of degree n has n complex solutions, roots, or zeros.

If x=a is a solution or root of a polynomial equation f\left(x\right)=0, then:

  • f\left(a\right)=0

  • x=a is a zero of the polynomial f\left(x\right)

  • f\left(x\right) has an x-intercept at \left(a,0\right)

  • \left(x-a\right) is a factor of f\left(x\right)

The multiplicity of a zero is the number of times a zero is repeated. This can be found by the exponent of its corresponding factor in the function. When graphed, the multiplicities appear as follows:

  • Multiplicity of 1 crosses through the x-axis

  • Even multiplicity is tangent to the x-axis

  • Odd multiplicity greater than 1 crosses through the x-axis with a point of inflection

Imaginary roots come in complex conjugate pairs like 7i and -7i.

Outcomes

A2.EI.6

The student will represent, solve, and interpret the solution to a polynomial equation.

A2.EI.6a

Determine a factored form of a polynomial equation, of degree three or higher, given its zeros or the x-intercepts of the graph of its related function.

A2.EI.6b

Determine the number and type of solutions (real or imaginary) of a polynomial equation of degree three or higher.

A2.F.2

The student will investigate and analyze characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions algebraically and graphically.

A2.F.2a

Determine and identify the domain, range, zeros, and intercepts of a function presented algebraically or graphically, including graphs with discontinuities.

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