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4. Polynomial Functions
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United States of America
Grades 9-12

4.02 Zeros and factors

Lesson
Practice

Introduction

In lesson  3.06 Solving polynomial equations  , we solved polynomials by factoring and using the zero product property. We also used identities from lesson  3.03 Polynomial identities  to solve nonlinear factors. If a quadratic factor could not be factored further, then we solved it using the quadratic formula. This lesson will connect these solutions to the graph of the polynomial function.

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 complex roots of any polynomial is equal to the degree of the polynomial. Remember that complex roots refer to real and imaginary roots. The real zeros of a function will be the x-intercepts of its graph.

Recall the factor theorem says if x=a is a root of 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
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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. A root with an odd multiplicity greater than 1 crosses through the x-axis with a point of inflection. 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.

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

  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 would need to know the roots, their multiplicities, the degree of the function, and another point on the graph in order 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

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.

Example 2

A polynomial f\left(x\right)=x^5+7x^4+17x^3+47x^2+72x-144 has zeros at x=-4 and x=-3i.

a

Determine the remaining zeros of f\left(x\right).

Worked Solution
Create a strategy

By the fundamental theorem of algebra, this function has 5 complex roots since the degree is 5. We were given 2 of the roots which means 3 still remain.

Since this polynomial has real coefficients, we can apply the complex conjugates theorem. We can then create factors for the known roots and use polynomial division to factor f\left(x\right) and find the remaining roots of the function.

Apply the idea

If a number k is a zero of f\left(x\right), then f\left(k\right)=0. It follows that x-k is a factor by the factor theorem. So, we have \left(x+4\right) and \left(x+3i\right) as factors of f\left(x\right).

Since this polynomial has real coefficients, we can apply the complex conjugates theorem which says if -3i is a root of the polynomial, then 3i must also be a root of the polynomial.

Multiplying together the factors we know, we get

\displaystyle \left(x+4\right)\left(x+3i\right)\left(x-3i\right)\displaystyle =\displaystyle \left(x+4\right)\left(x^2+9\right)Difference of squares identity
\displaystyle =\displaystyle x^3+4x^2+9x+36Distributive property

Next, we can use polynomial division to find P\left(x\right)\div \left(x^3+4x^2+9x+36\right) and factor the result to find the remaining zeros.

A figure showing the polynomial long division for x to the fifth power plus 7 x to the fourth power plus 17 x cubed plus 47 x squared plus 72 x minus 144 divided by x cubed plus 4 x squared plus 9 x plus 36. Speak to your teacher for more information.

The result, x^2+3x-4, can be factored further: x^2+3x-4=\left(x+4\right)\left(x-1\right) Setting these factors to zero and solving, we find the remaining roots are x=-4 and x=1, and the root we found earlier at x=3i.

Reflect and check

The zeros given to use were x=-4 and x=-3i, and the zeros we found were x=3i, x=-4, and x=1. We can check whether the zeros of f\left(x\right) are correct by multiplying their corresponding factors.

\displaystyle f\left(x\right)\displaystyle =\displaystyle \left(x-1\right)\left(x+4\right)\left(x+4\right)\left(x+3i\right)\left(x-3i\right)Function expressed by its factors
\displaystyle =\displaystyle \left(x-1\right)\left(x+4\right)^2\left(x^2+9\right)Difference of squares identity
\displaystyle =\displaystyle \left(x-1\right)\left(x^2+8a+16\right)\left(x^2+9\right)Perfect square trinomial identity
\displaystyle =\displaystyle \left(x^2+8a+16\right)\left(x^3-x^2+9x-9\right)Distribute \left(x-1\right)\left(x^2+9\right)
\displaystyle =\displaystyle x^5+7x^4+17x^3+47x^2+72x-144Distributive property

This shows that the zeros we obtained are correct.

b

State the multiplicities of the zeros.

Worked Solution
Create a strategy

The full factorization of the polynomial is f\left(x\right)=\left(x-1\right)\left(x+4\right)\left(x+4\right)\left(x+3i\right)\left(x-3i\right) which can be simplified to f\left(x\right)=\left(x-1\right)\left(x+4\right)^2\left(x+3i\right)\left(x-3i\right). To determine the multiplicities of each zero, we determine the number of times that the corresponding factor of each zero occurs in the function.

Apply the idea

From the factored form of the polynomial, we see

  • 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
Reflect and check

There are 5 roots which confirms the number of roots expected from the fundamental theorem of algebra.

Example 3

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

-5
-4
-3
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-1
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-40
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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=-\frac{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=-\frac{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 4

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

  • Degree of 3

  • Zeros include x=3 and x=-\sqrt{2}

  • Rational 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=-\sqrt{2}. The latter is an irrational root, so we need to determine if we can use the irrational conjugates theorem by looking at the information given about the coefficients. The coefficients must be rational in order to use the theorem.

Apply the idea

Since x=-\sqrt{2} is a zero and the polynomial has rational coefficients, then x=\sqrt{2} must also be a zero by the irrational conjugates theorem. From these, we get the expression \left(x-3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\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-\sqrt{2}\right)\left(x+\sqrt{2}\right)Equation of the function
\displaystyle 12\displaystyle =\displaystyle a\left(0-3\right)\left(0-\sqrt{2}\right)\left(0+\sqrt{2}\right)Substitute values from the y-intercept
\displaystyle 12\displaystyle =\displaystyle a\left(-3\right)\left(-\sqrt{2}\right)\left(\sqrt{2}\right)Evaluate the addition
\displaystyle 12\displaystyle =\displaystyle 6aEvaluate the multiplication
\displaystyle 2\displaystyle =\displaystyle aDivide both sides by 2

Therefore,f\left(x\right)=2\left(x-3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right) is the polynomial function.

Reflect and check

If the information did not tell us that the coefficients were rational, then we could not have used the irrational conjugates theorem. This theorem can only be used if the coefficients are rational because a polynomial with irrational coefficients can have only one irrational root.

Idea summary

The fundamental theorem of algebra says the number of complex zeros of any polynomial is equal to the degree of the polynomial. These include both real and imaginary zeros. The real zeros of a function will be the x-intercepts of its graph.

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

Outcomes

N.CN.C.9 (+)

Know the fundamental theorem of algebra; show that it is true for quadratic polynomials.

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.B.3

Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.

F.BF.A.1

Write a function that describes a relationship between two quantities.

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