Main content

### Course: Algebra (all content) > Unit 10

Lesson 37: Symmetry of polynomial functions# 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 $y$ -axis.

**even function**if its graph is symmetric with respect to theAlgebraically, $f$ is an even function if $f(-x)=f(x)$ 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 $f(-x)=-f(x)$ 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 $f(x)=a{x}^{n}$ where $a$ is a real number and $n$ is an integer greater than or equal to $0$ .

**monomial**is a one-termed polynomial. Monomials have the formIn 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* , then we know$f(x)$ is even.$f$ - If
$f(-x)$ *is the opposite of* , then we know$f(x)$ is odd.$f$ - Otherwise, it is neither even nor odd.

As a first example, let's determine whether $f(x)=4{x}^{3}$ is even, odd, or neither.

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.

### Concluding the investigation

From the above problems, we see that if $f$ is a monomial function of $f$ is an $f$ is a monomial function of $f$ is an

*even*degree, then function*even function*. Similarly, if*odd*degree, then function*odd function.*Even Function | Odd Function | |
---|---|---|

Examples | ||

In general |

This is because $(-x{)}^{n}={x}^{n}$ when $n$ is even and $(-x{)}^{n}=-{x}^{n}$ 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: $f(x)=2{x}^{4}-3{x}^{2}-5$

To determine whether $f$ is even, odd, or neither, we find $f(-x)$ .

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: $g(x)=5{x}^{7}-3{x}^{3}+x$

Again, we start by finding $g(-x)$ .

At this point, notice that each term in $g(-x)$ is the $g(x)$ . In other words, $g(-x)=-g(x)$ , and so $g$ is an odd function.

*opposite*of each term inNote that all the terms of $g$ are of an odd degree.

### Example 3: $h(x)=2{x}^{4}-7{x}^{3}$

Let's find $h(-x)$ .

*opposite*of

Mathematically, $h(-x)\ne h(x)$ and $h(-x)\ne -h(x)$ , 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.

General rule | Example polynomial | |
---|---|---|

Even | A polynomial is even if each term is an even function. | |

Odd | A polynomial is odd if each term is an odd function. | |

Neither | A polynomial is neither even nor odd if it is made up of both even and odd functions. |

### Check your understanding

## Want to join the conversation?

- Is a constant considered an "even" term?(44 votes)
- 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.)(75 votes)

- 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?(5 votes)
- 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.(12 votes)

- Are polynomial functions only even and odd? Or are exponential, logarithmic, etc. as well?(6 votes)
- 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.(7 votes)

- Does anybody know how this would apply to:

f(x)=SQRT(x)

Thanks!(5 votes)- 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.(6 votes)

- 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)
- 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!(8 votes)

- 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?(1 vote)
- 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.(9 votes)

- Can terms be functions?(2 votes)
- Yes, terms can be anything you can add. That includes functions, vectors, matrices, and numbers.(4 votes)

- why does (-x)^2=x^2 in the explanation of the 2nd question?(1 vote)
- (-x)^2 = (-x)(-x)

a negative times a negative = a positive

So, the final results is +x^2

Hope this helps.(5 votes)

- This sort of investigation does not fully determine anything... can anyone prove the even/odd/neither pattern Khan describes?(1 vote)
- 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.(5 votes)

- 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)