5.3 Polynomial Functions

Learning Objectives

By the end of this section, you will be able to:

• identify polynomial functions and their important features: degree, leading coefficient, roots, end behavior
• write polynomial functions in standard form
• find real roots in easy factoring situations
• identify the degree of a polynomial function based on its graph and understand how features of the graph relate to the function

In the book Watership Down, the rabbits can't really count. In their rabbit language, they have the counting words, "one, two, three, four.......many." In our survey of polynomial functions, we've done degree-one (linear) and degree-two (quadratic) polynomials, and now we're going to open it up to general degree-many. Oh, I mean degree-\(n\).

Polynomial Functions

A polynomial function is a function that can be written in the form

\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \notag \]

where \(n\) is a nonnegative integer, \(a_n \neq 0\), and \( a_0, a_1, ..., a_n\) are constants. Fun facts about polynomial functions:

• The highest appearing power \(n\) is called the degree.
• These functions always have all real numbers as their domain.
• The range of a polynomial function depends on whether its degree is even or odd and the sign of its leading coefficient. We'll talk about this and end behavior later.
• The \(x\)-intercepts are called the roots or zeros of the polynomial function. The maximum number of real roots for a given degree-\(n\) polynomial, counting with multiplicity, is \(n\).
• The graphs of polynomials are always nicely smoothly curved, with no skips or jumps or corners.

Let's be structured about how we look at example graphs. A lot can be told about the shape of a function's graph by knowing the degree and sign of the leading coefficient of the polynomial, and vice versa. Carefully study the following table column-by-column and try to notice any patterns.

degree 2 (quadratic)	degree 3 (cubic)	degree 4 (quartic)	degree 5
\( f(x) = x^2 \)	\( f(x) = x^3 \)	\( f(x) = x^4 \)	\( f(x) = x^5 \)
\( f(x) =(x-3)(x+2) \)	\( f(x) = (x-1)(x+1)(x-2) \)	\( f(x) = x(x-2)(x-1)(x+2)\)	\( f(x) = x(x-1)(x-2)(x+1)(x+2) \)
\( f(x) = -(x-1)^2 \)	\( f(x) = -(x-2)^2(x+1) \)	\( f(x) = -(x+1)^2(x-1)^2 \)	\( f(x) = -x(x-2)^4 \)
\( f(x) = -x^2 + 3x - 3 \)	\( f(x) = -x^3 - 4x^2 - 1 \)	\( f(x) = x^4 - x^2 - x + 2 \)	\( f(x) = -x^5 + x^3 + 1 \)

Here are the takeaway observations:

• Monomials like \(f(x) = x^n\) for even power \(n\) kind of look like parabolas but flattening out more at the origin. If \(n\) is odd, they look like the basic cubic, but flattening out more at the origin.
• Even-degree polynomials always come in and go out in the same direction. Look at degree-two and degree-four. There may be some squiggles in the middle, but in the end they will either go up forever or down forever in both \(x\) directions. If the leading coefficient is positive, they will open upward. If the leading coefficient is negative, they will open downward.
• Odd-degree polynomials always come in from one direction and leave in the other. Look at degree-three and degree-five. There may be squiggles in the middle, but eventually they go down forever to the left and up forever to the right, or vice versa. If the leading coefficient is positive, the function generally travels uphill, going down forever as \(x \rightarrow -\infty\) and up forever as \(x \rightarrow \infty\). If the leading coefficient is negative, the function generally travels downhill, going up forever as \(x \rightarrow -\infty\) and down forever as \( x \rightarrow \infty\).
• As for the squiggles, a degree-\(n\) polynomial function can "turn around" (have a peak or a trough) up to \(n-1\) times. For example, the degree-four polynomials can't have more than three turnarounds.
• A degree-\(n\) polynomial function can have up to \(n\) distinct real roots. For example, look at the cubics column. The first one has a single root at \(0\) (and it has multiplicity 3, which we'll talk about in five seconds, just hang on). The second has three distinct roots. The third has two distinct roots, and the one at \(x = 2\) has multiplicity 2, causing that bouncing-off behavior. The fourth has only one real root again (and secretly a complex conjugate pair of roots, otherwise). A cubic can't cross the \(x\)-axis more than three times, so it can only have up to 3 distinct real roots.

Here's the end behavior observations organized into a table.

degree vs. sign of leading coefficient	\( a_n > 0 \)	\( a_n < 0 \)
\( n\) even	as \(x \rightarrow \infty\), \(f(x) \rightarrow \infty\) as \(x \rightarrow -\infty\), \(f(x) \rightarrow \infty\)	as \(x \rightarrow \infty\), \(f(x) \rightarrow -\infty\) as \(x \rightarrow -\infty\), \(f(x) \rightarrow -\infty\)
\( n\) odd	as \(x \rightarrow \infty\), \(f(x) \rightarrow \infty\) as \(x \rightarrow -\infty\), \(f(x) \rightarrow -\infty\)	as \(x \rightarrow \infty\), \(f(x) \rightarrow -\infty\) as \(x \rightarrow -\infty\), \(f(x) \rightarrow \infty\)

Example \(\PageIndex{1}\)

Tell whether the functions are polynomial functions, and if necessary, write them in standard form. If they are, identify the degree and the leading coefficient.

1. \( f(x) = x^{16} + x^{11} - x^3 + 1 \)
2. \( g(x) = x^{-2} + x^{-1} + x + x^2 \)
3. \( h(x) = -2(x+1)(x-2)^2 \)
4. \( p(x) = \sqrt{x^3 + x} \)
5. \( q(x) = 3x^5 + (x-1)^2 \)
6. \( r(x) = \dfrac{ x^3 - 1}{x^2 + 2} \)

Solution

1. Yes, the degree is 16 and the leading coefficient is a secret 1.
2. No, the negative powers of \(x\) are not allowed in a polynomial.
3. Yes, when expanded this becomes \( h(x) = -2x^3 + 6x^2 - 8 \), so the degree is 3 and the leading coefficient is \(-2\).
4. No, the square root is not allowed.
5. Yes, when expanded and simplified this becomes \(q(x) = 3x^5 + x^2 - 2x + 1 \), so the degree is 5 and the leading coefficient is 3.
6. No, this is two polynomials being divided.

Exercise \(\PageIndex{1}\)

Tell whether the functions are polynomial functions, and if necessary, write them in standard form.

1. \( f(x) = x^3 + x + \dfrac{1}{x} \)
2. \( g(x) = 1 - x - x^2 - x^3 - x^4 \)
3. \( h(x) = (x^2 + 1)^{\frac{3}{2} } \)
4. \( p(x) = x^2 - 7x^3 + x \)
5. \( q(x) = x^7(x-1)(x-2) \)
6. \( r(x) = \dfrac{x^5 + x + 2}{x^7 + 9} \)

Answer

1. No.
2. Yes, degree 4, leading coefficient \(-1\).
3. No.
4. Yes, in order, \( p(x) = -7x^3 + x^2 + x \), so degree 3 and leading coefficient \(-7\).
5. Yes, \( q(x) = x^9 - 3x^8 + 2x^7 \), so degree 9 and leading coefficient 1.
6. No.

Example \(\PageIndex{2}\)

For the functions in the previous example that turned out to be polynomials, describe the end behavior.

1. \( f(x) = x^{16} + x^{11} - x^3 + 1 \)
2. \( h(x) = -2(x+1)(x-2)^2 \)
3. \( q(x) = 3x^5 + (x-1)^2 \)

Solution

1. This has even degree and positive leading coefficient, so as \(x \rightarrow \infty\), \(f(x) \rightarrow \infty\), and as \(x \rightarrow -\infty\), \( f(x) \rightarrow \infty\).
2. This has odd degree and negative leading coefficient (try to see why, or look back at the expanded standard form we found), so as \(x \rightarrow \infty\), \(h(x) \rightarrow -\infty\), and as \(x \rightarrow -\infty\), \( h(x) \rightarrow \infty\).
3. This has odd degree and positive leading coefficient, so so as \(x \rightarrow \infty\), \(q(x) \rightarrow \infty\), and as \(x \rightarrow -\infty\), \( q(x) \rightarrow -\infty\).

Exercise \(\PageIndex{2}\)

Do the same for the polynomials in the previous exercise.

1. \( g(x) = 1 - x - x^2 - x^3 - x^4 \)
2. \( p(x) = x^2 - 7x^3 + x \)
3. \( q(x) = x^7(x-1)(x-2) \)

Answer

1. As \(x \rightarrow \infty\), \(g(x) \rightarrow -\infty\), and as \(x \rightarrow -\infty\), \( g(x) \rightarrow -\infty\).
2. As \(x \rightarrow \infty\), \(p(x) \rightarrow \infty\), and as \(x \rightarrow -\infty\), \( p(x) \rightarrow \infty\).
3. As \(x \rightarrow \infty\), \(q(x) \rightarrow \infty\), and as \(x \rightarrow -\infty\), \( q(x) \rightarrow -\infty\).

Roots and Multiplicity

Here's a big ol' key to relating the equation of a polynomial to its roots. Remember how we found roots of quadratic functions? We said, "If you want to be a root, you gotta be on the \(x\)-axis, which means you gotta have a \(y\)-coordinate of 0, which means the function value spit out needs to be 0." So we set the function equal to 0 and solved for \(x\), and generally the easiest way to do that was to factor the quadratic. Then if any factor is zero, the entire expression will be wiped out, so we were able to see immediately what values of \(x\) would result in a root. When we factor a polynomial, the factors that appear and the roots of the function are two sides of the same coin.

When factoring a polynomial, you will break it down into a set of factors that are either linear factors of the form \( (x-c) \), or irreducible quadratic factors of the form \((ax^2 + bx + c)\), where that quadratic does not have real roots; i.e. the discriminant \( b^2 - 4ac\) turned out to be negative. So finding all real roots of a polynomial is as simple as looking at the linear factors \( (x-c)\) and writing down a list of the \(c\)'s. Note: you may have quadratic factors that you can find ugly real roots for using the quadratic formula, which are not considered irreducible. Technically you would still be able to write linear factors for them, it's just gross like

\[ p(x) = \left( x + \frac{\sqrt{5}}{2} \right)\left( x - \frac{\sqrt{5}}{2} \right)(x-1). \notag \]

Now, I mentioned this in the quadratic functions section, but there is a concept called multiplicity, which we can be precise about now that we know the factor-root association fact above.

The real number \( c\) is a root (zero) of a polynomial function \( p(x) \) if and only (iff) the factor \( (x-c)\) appears in the factorization of \( p(x) \). The multiplicity of the root \(c\) is the number of times that factor is repeated in the factorization. So if

\[ p(x) = (\: \cdots \: ) \cdots (x-c)^n \cdots (\: \cdots\: ) \notag \]

then the multiplicity of root \(c\) is \(n\).

We'll learn some tactics for factoring higher-degree polynomials in the next section, but for now I'm hard at work in the kitchen cooking up these examples so that the factoring is manageable. Don't say I never gave you nothin'. And don't forget about the factoring by grouping technique!

Example \(\PageIndex{3}\)

Find all real roots of the polynomial functions, with multiplicities.

1. \( f(x) = x^3 - x^2 - 2x \)
2. \( g(x) = (x-3)(x^2-x-6) \)
3. \( p(x) = 2x^3 - 2x + x^2 - 1 \)
4. \( q(x) = x^3 + x \)
5. \( h(x) = x^5 \)

Solution

1. First, factor completely like so: \( f(x) = x(x^2 - x -2) = x(x-2)(x+1) \). Notice that the factor of \(x\) could be seen as \( (x-0)\), and the factor \( (x+1) \) should be seen as \(( x - (-1)) \). Then we can see that the \(c\)'s that appear in those factors are \( -1, 0,\) and \( 2\). Each of those factors appears only once, so the multiplicity of each root is 1.
2. This is already partially factored, so finish the job to get \( g(x) = (x-3)(x-3)(x+2) = (x-3)^2(x+2) \). We read the roots off as \(x = -2, 3 \), but we notice that there are two copies of the factor \( (x-3)\). The root \(x=3\) has multiplicity 2, and \(x = -2\) has multiplicity 1.
3. Factoring by grouping, we have \( p(x) = 2x(x^2 - 1) + (x^2 - 1) = (2x + 1)(x^2 - 1) = (2x + 1)(x+1)(x-1) \). If you're unsure what to do with that \( (2x+1)\), remember that what's going on behind the scenes of solving for roots is setting each factor equal to zero and solving for \(x\). So from \(2x +1 = 0 \) we get a root \( x = -\frac{1}{2} \). The other roots are \(x = -1, 1 \). All three have multiplicity 1 again.
4. Factoring out the \(x\) in common, we have \( q(x) = x(x^2 + 1) \). That \(x\) factor out front tells us a real root \(x = 0\), but the \((x^2 + 1)\) factor is an irreducible quadratic with no real roots (try setting it equal to zero—you will end up trying to square root a negative number). So the only real root we report is \(x = 0\), with multiplicity 1.
5. Remember that this can be thought of as \( h(x) = (x-0)^5 \), so we have a real root of \(x = 0\), with multiplicity 5.

Exercise \(\PageIndex{3}\)

Find all real roots of the polynomial functions, with multiplicities.

1. \( f(x) = (x-17)^4\)
2. \( g(x) = x^3 + 9x \)
3. \( h(x) = x( x+1 + 2x^2 + 2x) \)
4. \( p(x) = (x +8)(x^2 + 9x + 8) \)

Answer

1. \(x = 17 \), mult 4.
2. \( x = 0 \), mult 1 (the other factor is an irreducible quadratic).
3. \( x = -1, - \frac{1}{2}, 0 \), all mult 1 (try grouping).
4. \( x = -8 \) with mult 2, \(x = -1\) with mult 1.

The last thing I want to talk about in this section is how to recognize multiplicities from graphs. If you look back at the table of polynomial examples, you will notice a couple things.

• When a root has multiplicity 1, the graph passes straight through the \(x\)-axis at that point.
• When a root has multiplicity 2, the graph bounces off the \(x\)-axis at that point.
• When a root has multiplicity 3, the graph passes through the \(x\)-axis, but it hangs out for a second, flattening out a bit.
• When a root has multiplicity 4, the graph bounces off the \(x\)-axis, but it hangs out for a second, flattening out a bit.
• The pattern continues: pass through the \(x\)-axis at odd multiplicity roots, but don't cross over at even multiplicity roots, and the higher the multiplicity, the more flattening behavior.

Example \(\PageIndex{4}\)

Based on the graph given, determine a possible set of real roots with their multiplicities, write the factors that must appear in the factorization of the polynomial function, and use end behavior to determine the sign of the leading coefficient.

1.	2.	3.	4.

Solution

1. The graph bounces off at \(x = -1\), so that's a root with multiplicity 2. It passes straight through \(x = 4\), so that's a root with multiplicity 1. The factors \( (x+1)^2\) and \( (x-4)\) must appear. The graph comes in from below and leaves going up, so it is indeed a degree-3 polynomial with positive leading coefficient.
2. The graph passes straight through \(x = -3, -2, 1, 4 \), so those are all roots with multiplicity 1. The factors that must appear are \( (x+3)(x+2)(x-1)(x-4)\). This graph comes in from below and leaves going down, so it is indeed a degree-4 polynomial with negative leading coefficient.
3. The graph hangs out a bit at \(x = 3\) but ultimately doesn't pass through, so this looks like a root with multiplicity 4, so the factors must be \( (x-3)^4\). Again, coming in from below and leaving going down tells us it's a degree-4 polynomial with negative leading coefficient.
4. The graph passes straight through \(x = -2\) and \(x = 0\), roots with multiplicity 1. It hangs out a bit and then passes through at \(x = 2\), so that's a root with multiplicity 3. The factors are \( x(x+2)(x-2)^3 \), and this is a degree-5 polynomial with positive leading coefficient.

Now you try, and then hit the exercises!

Exercise \(\PageIndex{4}\)

Based on the graph given, determine a possible set of real roots with their multiplicities, write the factors that must appear in the factorization of the polynomial function, and use end behavior to determine the sign of the leading coefficient.

1.	2.	3.	4.

Answer

1. Roots: \(-4\) (mult 1), \(-2\) (mult 1), \(0\) (mult 2). Factors: \( x^2(x+4)(x+2)\). Degree-4 with positive leading coefficient.
2. Roots: \(0\) (mult 3), \(2\) (mult 1). Factors: \( x^3(x-2) \). Degree-4 with negative leading coefficient.
3. Roots: \( -2\) and \( 3\), both mult 1. Factors: \( (x+2)(x-3)\). Quadratic with positive leading coefficient.
4. Roots: \( -1\) (mult 4), \( 1\) (mult 1). Factors: \( (x+1)^4(x-1) \). Degree-5 with positive leading coefficient.