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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 parabola symmetric at x equals zero. on an x y coordinate plane. Its vertex is at (zero, zero). As x goes to negative infinity, the y value goes to infinity. As x goes to infinity, the y value goes to infinity.
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(x)=f(x) for all x.
An cubic function on an x y coordinate plane. Its middle point is at (zero, zero). As x goes to negative infinity, y goes to negative infinity. As x goes to infinity, y goes to infinity. The graph is concave down from the interval negative infinity to zero. The graph is concave up from the interval zero to infinity.
A function is an odd function if its graph is symmetric with respect to the origin.
Algebraically, f is an odd function if f(x)=f(x) 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(x)=axn 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(x):
  • If f(x) is the same as f(x), then we know f is even.
  • If f(x) is the opposite of f(x), then we know f is odd.
  • Otherwise, it is neither even nor odd.
As a first example, let's determine whether f(x)=4x3 is even, odd, or neither.
f(x)=4(x)3=4(x3)(x)3=x3=4x3Simplify=f(x)Since f(x)=4x3
Here f(x)=f(x), 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(x)=3x2 even, odd, or neither?
Choose 1 answer:

2) Is h(x)=2x5 even, odd, or neither?
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(x)=3x2h(x)=2x5
In generalf(x)=axn where n is evenf(x)=axn where n is odd
This is because (x)n=xn when n is even and (x)n=xn 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(x)=2x43x25

To determine whether f is even, odd, or neither, we find f(x).
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
Since f(x)=f(x), function f is an even function.
Note that all the terms of f are of an even degree.

Example 2: g(x)=5x73x3+x

Again, we start by finding g(x).
g(x)=5(x)73(x)3+(x)=5(x7)3(x3)+(x)(x)n=xn when n is odd=5x7+3x3xSimplify
At this point, notice that each term in g(x) is the opposite of each term in g(x). In other words, g(x)=g(x), and so g is an odd function.
Note that all the terms of g are of an odd degree.

Example 3: h(x)=2x47x3

Let's find h(x).
h(x)=2(x)47(x)3=2(x4)7(x3)(x)4=x4 and (x)3=x3=2x4+7x3Simplify
2x4+7x3 is not the same as h(x) nor is it the opposite of h(x).
Mathematically, h(x)h(x) and h(x)h(x), 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.
xGeneral ruleExample polynomial
EvenA polynomial is even if each term is an even function.f(x)=2x43x25
OddA polynomial is odd if each term is an odd function.g(x)=5x73x3+x
NeitherA polynomial is neither even nor odd if it is made up of both even and odd functions.h(x)=2x47x3

Check your understanding

3) Is f(x)=3x47x2+5 even, odd, or neither?
Choose 1 answer:

4) Is g(x)=8x76x3+x2 even, odd, or neither?
Choose 1 answer:

5) Is h(x)=10x5+2x3x even, odd, or neither?
Choose 1 answer:

Want to join the conversation?

  • aqualine ultimate style avatar for user Tony Fiore
    Is a constant considered an "even" term?
    (36 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?
    (3 votes)
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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.
      (8 votes)
  • mr pink orange style avatar for user freddyfazbearspizza
    Are polynomial functions only even and odd? Or are exponential, logarithmic, etc. as well?
    (4 votes)
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    • stelly blue style avatar for user Kim Seidel
      Polynomials functions may or may not be even or odd. As soon as you shift a graph left/right or up/down, you may lose any y-axis or origin symmetry that may have existed. For example: y=x^2 has y-axis symmetry and is an even function. y=(x+1)^2 no longer has y-axis symmetry and is no longer an even function.

      Hope this helps.
      (3 votes)
  • blobby green style avatar for user John Murphy
    Now that I know the "general rule" I know how to correctly solve these type of problems. But just abstractly solving without using a real number for x, the test still confuses me. I think my confusion has to do with f(-x) and -f(-x). So the question is who do I put that idea more concretely in my mind?
    (2 votes)
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    • male robot donald style avatar for user Venkata
      So look, the symmetry of any function basically depends on f(-x), where x is some real number within the domain. So, the article says that if f(-x) is the same as the function (i.e. f(-x) = f(x)), the function is even. Conversely, if f(-x) is the opposite of the function (i.e. f(-x) = -f(x), NOT -(f(-x))), the function is odd.

      I'm surprised they didn't use a proper number example here, but here you go:

      So, let's use the same example given. So, let y = 5x^7−3x^3+x. From the proof in Example 3, we know this function is odd. Let's test it.

      Let's first find f(2). So, we have y = 5(2)^(7) - 3(2)^(3) + 2. This gives us 640 - 24 + 2 = 618.

      Now, let's find f(-2). So, we have y = 5(-2)^(7) - 3(-2)^(3) + (-2). This gives us -640 + 24 - 2 = -618.

      Now, see that f(x) isn't equal to f(-x). So, the function isn't even. Now, let's find -f(2). We know f(2) = 618. So, -f(2) is simply -618, which is exactly what f(-2) is. Hence, f(-2) = -f(2) and thus, the function is odd.

      Hope this helped!
      (6 votes)
  • aqualine ultimate style avatar for user Brigette Stoddard
    Does anybody know how this would apply to:
    f(x)=SQRT(x)

    Thanks!
    (3 votes)
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    • piceratops ultimate style avatar for user Jimmy
      since x is under a square root sign, x can only be positive, eliminating any chance of x being negetive, so there is no -x in our graph, so f(-x) does not exist. Therefore, f(x) is not an even or an odd function. Hope this cleared things up.
      (3 votes)
  • 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.
      (7 votes)
  • piceratops ultimate 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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  • aqualine ultimate style avatar for user Gabyscience
    This sort of investigation does not fully determine anything... can anyone prove the even/odd/neither pattern Khan describes?
    (1 vote)
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    • blobby blue style avatar for user joshua
      If you substitute -x and you still get the same result as you substitute x, it means that the function will output the same results for x = a and x = -a. This explains the even pattern.

      For odd function, you know that if you separate the graph at the origin (It must pass through it), both the x and y will have opposite signs. Thus if you use the method same as even but get a negative of the initial function, you know that it's odd.

      Well neither is, neither.
      (4 votes)
  • aqualine ultimate style avatar for user Simum
    Can terms be functions?
    (1 vote)
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  • leaf blue style avatar for user bryancatacata2
    As far as identifying even/odd functions graphically, is it true that an even function will have an even amount of points of inflection, and vice versa for odd? This would be assuming that you only search within a domain with "edges" that are equally distanced from the y-axis. I assume that this is correct because the number of inflection points is the same as the number of the exponent of the highest degree, so odd inflections always corresponds to odd exponent?
    (2 votes)
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