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Symmetry of polynomials

Learn how to determine if a polynomial function is even, odd, or neither.

What you should be familiar with before taking this lesson

A function is an even function if its graph is symmetric with respect to the y-axis.
Algebraically, f is an even function if f, left parenthesis, minus, x, right parenthesis, equals, f, left parenthesis, x, right parenthesis for all x.
A function is an odd function if its graph is symmetric with respect to the origin.
Algebraically, f is an odd function if f, left parenthesis, minus, x, right parenthesis, equals, minus, f, left parenthesis, x, right parenthesis for all x.
If this is new to you, we recommend that you check out our intro to symmetry of functions.

What you will learn in this lesson

You will learn how to determine whether a polynomial is even, odd, or neither, based on the polynomial's equation.

Investigation: Symmetry of monomials

A monomial is a one-termed polynomial. Monomials have the form f, left parenthesis, x, right parenthesis, equals, a, x, start superscript, n, end superscript where a is a real number and n is an integer greater than or equal to 0.
In this investigation, we will analyze the symmetry of several monomials to see if we can come up with general conditions for a monomial to be even or odd.
In general, to determine whether a function f is even, odd, or neither even nor odd, we analyze the expression for f, left parenthesis, minus, x, right parenthesis:
  • If f, left parenthesis, minus, x, right parenthesis is the same as f, left parenthesis, x, right parenthesis, then we know f is even.
  • If f, left parenthesis, minus, x, right parenthesis is the opposite of f, left parenthesis, x, right parenthesis, then we know f is odd.
  • Otherwise, it is neither even nor odd.
As a first example, let's determine whether f, left parenthesis, x, right parenthesis, equals, 4, x, cubed is even, odd, or neither.
f(x)=4(x)3=4(x3)(x)3=x3=4x3Simplify=f(x)Since f(x)=4x3\begin{aligned}f(\blueD{-x})&=4(\blueD{-x})^3\\ \\ &=4(-x^3)&&\small{\gray{(-x)^3=-x^3}}\\ \\ &=-4x^3&&\small{\gray{\text{Simplify}}}\\ \\ &=-f(x)&&\small{\gray{\text{Since $f(x)=4x^3$}}}\\ \end{aligned}
Here f, left parenthesis, minus, x, right parenthesis, equals, minus, f, left parenthesis, x, right parenthesis, and so function f is an odd function.
Now try some examples on your own to see if you can find a pattern.
1) Is g, left parenthesis, x, right parenthesis, equals, 3, x, squared even, odd, or neither?
Choose 1 answer:
Choose 1 answer:

2) Is h, left parenthesis, x, right parenthesis, equals, minus, 2, x, start superscript, 5, end superscript even, odd, or neither?
Choose 1 answer:
Choose 1 answer:

Concluding the investigation

From the above problems, we see that if f is a monomial function of even degree, then function f is an even function. Similarly, if f is a monomial function of odd degree, then function f is an odd function.
Even FunctionOdd Function
Examples g, left parenthesis, x, right parenthesis, equals, 3, x, start superscript, start color #aa87ff, 2, end color #aa87ff, end superscripth, left parenthesis, x, right parenthesis, equals, minus, 2, x, start superscript, start color #1fab54, 5, end color #1fab54, end superscript
In generalf, left parenthesis, x, right parenthesis, equals, a, x, start superscript, start color #aa87ff, n, end color #aa87ff, end superscript where n is start color #aa87ff, start text, e, v, e, n, end text, end color #aa87fff, left parenthesis, x, right parenthesis, equals, a, x, start superscript, start color #1fab54, n, end color #1fab54, end superscript where n is start color #1fab54, start text, o, d, d, end text, end color #1fab54
This is because left parenthesis, minus, x, right parenthesis, start superscript, n, end superscript, equals, x, start superscript, n, end superscript when n is even and left parenthesis, minus, x, right parenthesis, start superscript, n, end superscript, equals, minus, x, start superscript, n, end superscript when n is odd.
This is probably the reason why even and odd functions were named as such in the first place!

Investigation: Symmetry of polynomials

In this investigation, we will examine the symmetry of polynomials with more than one term.

Example 1: f, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 3, x, squared, minus, 5

To determine whether f is even, odd, or neither, we find f, left parenthesis, minus, x, right parenthesis.
f(x)=2(x)43(x)25=2(x4)3(x2)5(x)n=xn when n is even=2x43x25Simplify=f(x)Since f(x)=2x43x25\begin{aligned}f(\blueD{-x})&=2(\blueD{-x})^4-3(\blueD{-x})^2-5\\ \\ &=2(x^4)-3(x^2)-5&&\small{\gray{(-x)^n=x^n\text{ when $n$ is even}}}\\ \\ &=2x^4-3x^2-5&&\small{\gray{\text{Simplify}}}\\\\ &=f(x)&&\small{\gray{\text{Since } f(x)=2x^4-3x^2-5}} \end{aligned}
Since f, left parenthesis, minus, x, right parenthesis, equals, f, left parenthesis, x, right parenthesis, function f is an even function.
Note that all the terms of f are of an even degree.

Example 2: g, left parenthesis, x, right parenthesis, equals, 5, x, start superscript, 7, end superscript, minus, 3, x, cubed, plus, x

Again, we start by finding g, left parenthesis, minus, x, right parenthesis.
g(x)=5(x)73(x)3+(x)=5(x7)3(x3)+(x)(x)n=xn when n is odd=5x7+3x3xSimplify\begin{aligned}g(\blueD{-x})&=5(\blueD{-x})^7-3(\blueD{-x})^3+(\blueD{-x})\\ \\ &=5(-x^7)-3(-x^3)+(-x)&&\small{\gray{(-x)^n=-x^n\text{ when $n$ is odd}}}\\ \\ &=-5x^7+3x^3-x&&\small{\gray{\text{Simplify}}}\\ \end{aligned}
At this point, notice that each term in g, left parenthesis, minus, x, right parenthesis is the opposite of each term in g, left parenthesis, x, right parenthesis. In other words, g, left parenthesis, minus, x, right parenthesis, equals, minus, g, left parenthesis, x, right parenthesis, and so g is an odd function.
Note that all the terms of g are of an odd degree.

Example 3: h, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 7, x, cubed

Let's find h, left parenthesis, minus, x, right parenthesis.
h(x)=2(x)47(x)3=2(x4)7(x3)(x)4=x4 and (x)3=x3=2x4+7x3Simplify\begin{aligned}h(\blueD{-x})&=2(\blueD{-x})^4-7(\blueD{-x})^3\\ \\ &=2(x^4)-7(-x^3)&&\small{\gray{(-x)^4=x^4\text{ and } (-x)^3=-x^3}}\\ \\ &=2x^4+7x^3&&\small{\gray{\text{Simplify}}}\\\\ \end{aligned}
2, x, start superscript, 4, end superscript, plus, 7, x, cubed is not the same as h, left parenthesis, x, right parenthesis nor is it the opposite of h, left parenthesis, x, right parenthesis.
Mathematically, h, left parenthesis, minus, x, right parenthesis, does not equal, h, left parenthesis, x, right parenthesis and h, left parenthesis, minus, x, right parenthesis, does not equal, minus, h, left parenthesis, x, right parenthesis, and so h is neither even nor odd.
Note that h has one even-degree term and one odd-degree term.

Concluding the investigation

In general, we can determine whether a polynomial is even, odd, or neither by examining each individual term.
empty spaceGeneral ruleExample polynomial
EvenA polynomial is even if each term is an even function.f, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 3, x, squared, minus, 5
OddA polynomial is odd if each term is an odd function.g, left parenthesis, x, right parenthesis, equals, 5, x, start superscript, 7, end superscript, minus, 3, x, cubed, plus, x
NeitherA polynomial is neither even nor odd if it is made up of both even and odd functions.h, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 7, x, cubed

Check your understanding

3) Is f, left parenthesis, x, right parenthesis, equals, minus, 3, x, start superscript, 4, end superscript, minus, 7, x, squared, plus, 5 even, odd, or neither?
Choose 1 answer:
Choose 1 answer:

4) Is g, left parenthesis, x, right parenthesis, equals, 8, x, start superscript, 7, end superscript, minus, 6, x, cubed, plus, x, squared even, odd, or neither?
Choose 1 answer:
Choose 1 answer:

5) Is h, left parenthesis, x, right parenthesis, equals, 10, x, start superscript, 5, end superscript, plus, 2, x, cubed, minus, x even, odd, or neither?
Choose 1 answer:
Choose 1 answer:

Want to join the conversation?

  • aqualine ultimate style avatar for user Tony Fiore
    Is a constant considered an "even" term?
    (35 votes)
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  • male robot hal style avatar for user Hesham
    Why would you use the F(x)=F(-x)/-F(x) rule when you can simply check whether the exponents are odd or even? Why check for points on the graph when you clearly know that if it's not symmetrical, it's not even? Is it just to figure out whether a graph is perfectly even or perfectly odd?
    (1 vote)
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    • mr pink red style avatar for user andrewp18
      There is no such thing as "perfectly even or odd". There is only even or odd or neither.
      Not all functions we wish to classify are polynomials. For instance, how would you tell me if the sine function was even or odd or neither? Good observation though – all polynomial functions only have parity if all exponents have the same parity.
      (5 votes)
  • piceratops tree style avatar for user Quinny The Pooh
    why does (-x)^2=x^2 in the explanation of the 2nd question?
    (1 vote)
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  • blobby green style avatar for user shatw
    i want to know about absolute value if it's even or odd
    (1 vote)
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  • blobby green style avatar for user ziad43636
    What do I do if the function is in a factored form and its degree is, so high that it will be hard to deduce its original form?
    (0 votes)
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    • blobby green style avatar for user kea241199
      You could just get the explicit definition of f(-x) and/or -f(-x) and see if they are equal to the definition of f(x) to determine if f(x) is odd, even, or neither.

      EDIT:
      I've thought about your question for a while. Another scenario would be is what if there were literally so many factors that replacing x's with -x's, negating, and then simplifying the whole expression was too cumbersome? So, I've devised a method that comes directly from the basic arithmetic properties of odd and even functions.

      Let P(x) = a ⋅ x₁ ⋅ x₂ ⋅ ... ⋅ xₙ ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ) where P(x) is a polynomial in factored form and each factor is linear—this is possible for any polynomial according to The Linear Factorization Theorem.

      a is an even function because a = a ⋅ x⁰. Factors of the form xₖ are odd functions because xₖ = xₖ¹. And factors of the form (x - cⱼ) are neither odd nor even (let's call them noden for short) since x - cⱼ = x¹ - cⱼ ⋅ x⁰.

      Here are some properties of odd, even, and noden functions (each function is strictly of that parity). For this section, I'm going to use even to denote an even function, odd to denote an odd function, and noden to denote a function that is neither odd nor even—all of which are polynomials. When I use any of those terms multiple times, they do not necessarily refer to the same function.

      even ⋅ even = even
      odd ⋅ even = odd
      noden ⋅ even = noden

      odd ⋅ odd = even
      Or more generally,
      odd₁ ⋅ odd₂ ⋅ ... ⋅ oddₙ = even ; iff. n is even
      odd₁ ⋅ odd₂ ⋅ ... ⋅ oddₙ = odd ; iff. n is odd

      odd ⋅ noden = noden

      noden ⋅ noden = even ; if both nodens are linear of the form (x - c) and are conjugates of each other. The simplest example is (x + a) and (x - a), they are conjugates of each other (I'll refer to them as a conjugate pair). Their product is x² - a², which is an even function. The product of any conjugate pair is always an even function.

      From those properties, we can see that multiplying by an even function doesn't affect the parity of the product. When we only care about parity, we can ignore the even functions/factors in a factored polynomial. I'm going to use p as a variable in the context of algorithms and := as an assignment operator. Let's assign p as P(x) without a, we can ignore it for the sake of the overall parity of P(x).

      p := x₁ ⋅ x₂ ⋅ ... ⋅ xₙ ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)

      p has the same parity as P(x).

      Now, let's determine what n is, which tells us how many factors of the form xₖ there are. Remember, factors of the form xₖ are odd functions, so when there are an even number of them, their product is an even function, otherwise (if the n is odd), the product is an odd function.

      If n is even,
      p := (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)
      otherwise, if n is odd,
      p := x ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cₘ)

      p and P(x) still have the same parity.

      Now, let's remove all the conjugate pairs. Remember that the product of any conjugate pair is always an even function. I'm going to denote that operation as a method, RemoveConjugatePairs.

      p := RemoveConjugatePairs(p)

      p and P(x) still have the same parity.

      if p = 1, P(x) is an even function.
      ; we removed all the even functions, and we are left with 1, that means P(x) was an even function all along. Another way interpret this is that 1 is an even function, so P(x) is an even function.

      if p = (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ), P(x) is a noden function.
      ; the product of noden functions of the form (x - c) is a noden function if there exists a noden factor which is not part of a conjugate pair—all instances of conjugate pairs have already been removed of course.

      if p = x, then P(x) is an odd function.
      ; x is an odd function so P(x) is an odd function.

      if p = x ⋅ (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ), P(x) is a noden function.
      ; we already know that (x - c₁) ⋅ (x - c₂) ⋅ ... ⋅ (x - cⱼ) is a noden function so if we multiply that by an odd function, the product will be a noden function.

      So now we know the parity of P(x). Hopefully you'll have less of a headache determining the parity of extremely large polynomials in factored form.
      (3 votes)
  • blobby green style avatar for user ChibwabwaK
    i want to know about absolute value if it´s even or odd
    (1 vote)
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  • duskpin ultimate style avatar for user Isabel
    Is every function an odd function if the degree is odd and the function goes through the origin?
    (1 vote)
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  • aqualine ultimate style avatar for user Simum
    Can terms be functions?
    (0 votes)
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  • blobby green style avatar for user mc40387
    I am not sure i understand how to determine the odd or even graphs.
    (0 votes)
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  • leafers seed style avatar for user Collin
    I don't think this was said explicitly: Are even-ness and odd-ness the only types of symmetry? So when someone says that a function is "symmetric", are they saying that it is guaranteed to be even or odd?
    (0 votes)
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    • leaf green style avatar for user kubleeka
      A symmetry is any way we can move a function around, and have the end result look the same as the beginning.

      You may be able to come up with other types of symmetry (for example, y=0 is symmetric over the x-axis), but when someone just says 'this function is symmetrical' with no context, they usually mean it is even.
      (1 vote)