Is a constant considered an "even" term?

Yes. Constant functions are symmetric about the y axis and are even functions. (You can think of the power of the x variable as zero, which is an even number, to help you remember this.)

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?

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.

why does (-x)^2=x^2 in the explanation of the 2nd question?

(-x)^2 = (-x)(-x) a negative times a negative = a positive So, the final results is +x^2 Hope this helps.

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?

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.

Can terms be functions?

Yes, terms can be anything you can add. That includes functions, vectors, matrices, and numbers.

Main content

Algebra 2

Course: Algebra 2 > Unit 9

Lesson 3: Symmetry of functions

Symmetry of polynomials

CCSS.Math:

HSF.BF.B.3

Google Classroom

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.

$\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.

[I want to see the graph to check if f is really odd.]

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?

[I need help!]

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

[I need help!]

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 Function	Odd Function
Examples	g, left parenthesis, x, right parenthesis, equals, 3, x, start superscript, start color #aa87ff, 2, end color #aa87ff, end superscript	h, left parenthesis, x, right parenthesis, equals, minus, 2, x, start superscript, start color #1fab54, 5, end color #1fab54, end superscript
In general	f, 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 #aa87ff	f, 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.

\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.

\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.

\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 space	General rule	Example polynomial
Even	A 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
Odd	A 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
Neither	A 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?

[I need help!]

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?

[I need help!]

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?

[I need help!]

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Tony Fiore
Posted 7 years ago. Direct link to Tony Fiore's post “Is a constant considered ...”
Is a constant considered an "even" term?
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(35 votes)
Answer
- Paul Miller
  Posted 7 years ago. Direct link to Paul Miller's post “Yes. Constant functions ...”
  Yes. Constant functions are symmetric about the y axis and are even functions. (You can think of the power of the x variable as zero, which is an even number, to help you remember this.)
  Comment on Paul Miller's post “Yes. Constant functions ...”
  (57 votes)
Hesham
Posted 6 years ago. Direct link to Hesham's post “Why would you use the F(x...”
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?
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(1 vote)
Answer
- andrewp18
  Posted 6 years ago. Direct link to andrewp18's post “There is no such thing as...”
  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.
  Comment on andrewp18's post “There is no such thing as...”
  (5 votes)
Quinny The Pooh
Posted 2 months ago. Direct link to Quinny The Pooh's post “why does (-x)^2=x^2 in th...”
why does (-x)^2=x^2 in the explanation of the 2nd question?
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(1 vote)
Answer
- Kim Seidel
  Posted 2 months ago. Direct link to Kim Seidel's post “(-x)^2 = (-x)(-x) a negat...”
  (-x)^2 = (-x)(-x)
  a negative times a negative = a positive
  So, the final results is +x^2
  
  Hope this helps.
  Comment on Kim Seidel's post “(-x)^2 = (-x)(-x) a negat...”
  (3 votes)
shatw
Posted 2 years ago. Direct link to shatw's post “i want to know about abso...”
i want to know about absolute value if it's even or odd
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(1 vote)
Answer
ziad43636
Posted a year ago. Direct link to ziad43636's post “What do I do if the funct...”
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?
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(0 votes)
Answer
- kea241199
  Posted a year ago. Direct link to kea241199's post “You could just get the ex...”
  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.
  Comment on kea241199's post “You could just get the ex...”
  (3 votes)
ChibwabwaK
Posted 2 years ago. Direct link to ChibwabwaK's post “i want to know about abso...”
i want to know about absolute value if it´s even or odd
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(1 vote)
Answer
- pi_is_3.14
  Posted 2 years ago. Direct link to pi_is_3.14's post “It is even since F(x) = F...”
  It is even since F(x) = F(-x)
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  (1 vote)
Isabel
Posted 3 years ago. Direct link to Isabel's post “Is every function an odd ...”
Is every function an odd function if the degree is odd and the function goes through the origin?
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- loumast17
  Posted 3 years ago. Direct link to loumast17's post “Not necessarily. somethi...”
  Not necessarily. something like x(x-1)(x-2) has a degree of 3 but doesn't fit the definition of an odd function.
  
  Also there are functions that are not polynomials that can be even or odd.
  Comment on loumast17's post “Not necessarily. somethi...”
  (1 vote)
Simum
Posted a year ago. Direct link to Simum's post “Can terms be functions?”
Can terms be functions?
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- kubleeka
  Posted a year ago. Direct link to kubleeka's post “Yes, terms can be anythin...”
  Yes, terms can be anything you can add. That includes functions, vectors, matrices, and numbers.
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  (2 votes)
mc40387
Posted 3 years ago. Direct link to mc40387's post “I am not sure i understan...”
I am not sure i understand how to determine the odd or even graphs.
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Answer
- obiwan kenobi
  Posted 3 years ago. Direct link to obiwan kenobi's post “Even graphs are symmetric...”
  Even graphs are symmetric over the y-axis. y=x^2 is a even graph because it is symmetric over the y-axis. Odd graphs are symmetric over the origin. y = x^3 is an odd graph because it is symmetric over the origin.
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  (1 vote)
Collin
Posted 4 years ago. Direct link to Collin's post “I don't think this was sa...”
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?
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- kubleeka
  Posted 4 years ago. Direct link to kubleeka's post “A symmetry is any way we ...”
  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.
  Comment on kubleeka's post “A symmetry is any way we ...”
  (1 vote)

Algebra 2

Course: Algebra 2 > Unit 9

What you should be familiar with before taking this lesson

What you will learn in this lesson

Investigation: Symmetry of monomials

Concluding the investigation

Investigation: Symmetry of polynomials

Example 1: f(x)=2x4−3x2−5f(x)=2x^4-3x^2-5f(x)=2x4−3x2−5f, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 3, x, squared, minus, 5

Example 2: g(x)=5x7−3x3+xg(x)=5x^7-3x^3+xg(x)=5x7−3x3+xg, left parenthesis, x, right parenthesis, equals, 5, x, start superscript, 7, end superscript, minus, 3, x, cubed, plus, x

Example 3: h(x)=2x4−7x3h(x)=2x^4-7x^3h(x)=2x4−7x3h, left parenthesis, x, right parenthesis, equals, 2, x, start superscript, 4, end superscript, minus, 7, x, cubed

Concluding the investigation

Check your understanding

Want to join the conversation?

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

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

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