3.3 - Real Zeros of Polynomial Functions

Long Division of Polynomials

You were taught long division of polynomials in Intermediate Algebra. Basically, the procedure is carried out like long division of real numbers. The procedure is explained in the textbook if you're not familiar with it.

One key point about division, and this works for real numbers as well as for polynomial division, needs to be pointed out. When you divide the dividend by the divisor, you get a quotient and a remainder. To check the problem, you multiply the divisor by the quotient and add the remainder to get the dividend. If the remainder is 0, then we say that the divisor divides evenly into the dividend.

Dividend / Divisor = Quotient + Remainder / Divisor

Dividend = Divisor * Quotient + Remainder

Like I said, the same thing can be done with polynomial functions.

f(x) = d(x) * q(x) + r(x)

Where f(x) is the polynomial function being divided into (dividend), d(x) is the polynomial function being divided by (divisor), q(x) is the polynomial function that is the quotient, and r(x) is the polynomial remainder function and will have degree less than the divisor.

If the remainder, r(x), is zero, then f(x) = d(x)*q(x). We have just factored the function f(x) into two factors, d(x) and q(x).

Remainder Theorem

When a polynomial function f is divided by x-k, the remainder r is f(k).

Okay, now in English. If you divide a polynomial by a linear factor, x-k, the remainder is the value you would get if you plugged x=k into the function and evaluated.

Now, tie that into what we just said above. If the remainder is zero, then you have successfully factored the polynomial. If the remainder when dividing by (x-k) is zero, then the function evaluated at x=k is zero and you have found a zero or root of the polynomial. Plus, you now have a factored polynomial (the quotient) which is one less degree than the original polynomial. If the quotient is down to a quadratic or linear factor, then you can solve and find the other solutions.

Synthetic Division

To divide a polynomial synthetically by x-k, perform the following steps.

Setup

1. Write k down, leave some space after it.
2. On the same line, right the coefficients of the polynomial function. Make sure you write the coefficients in order of decreasing power. Be sure to put a zero down if a power is missing. Place holders are very important
3. For now, leave a blank line. Draw the left and bottom portions of a box. The left portion goes between the k and the coefficients. The bottom portion goes under the blank line you left.

Synthetic Division

Once you have things set up, you can actually start to perform the synthetic division.

1. Bring the first coefficient down to the bottom row (below the line)
2. Multiply the number in the bottom row by the constant k, and write the product in the next column of the second row (above the line).
3. Add the numbers in the next column and write the total below the line.
4. Repeat steps 2 and 3 until all the columns are filled.

Interpreting the Results

1. The very last value is the remainder. If the remainder is zero, you have found a zero of the function.
2. The rest of the values are the coefficients of the quotient. Each term will be raised to the one less power than the original dividend. (If it was a fourth degree polynomial to start with, the quotient will be a third degree polynomial).

Warnings

You can only use synthetic division as described above to divide by x-k. That is, it must be a linear factor, and the leading coefficient must be a one.

There are similar ways to divide by a quadratic, cubic, etc, but for some reason, they aren't taught anymore (no, they won't die with me, I'm sure someone else knows them, too, but thanks for your concern).

Fundamental Theorem of Algebra

Every polynomial in one variable of degree n, n > 0, has at least one real or complex zero.

Corollary to the Fundamental Theorem of Algebra

Every polynomial in one variable of degree n, n > 0, has exactly n real or complex zeros.

Complex Roots

Complex solutions come in pairs. If (a+bi) is a solution, then its complex conjugate (a-bi) is also a solution.

Square Roots

Solutions involving square roots also come in pairs. If a+sqrt(b) is a solution, then its conjugate, a-sqrt(b) will also be a solution. The same is not necessarily true of other roots.

Descartes' Rule of Signs

This is not in your text!

• The maximum number of positive real roots can be found by counting the number of sign changes in f(x). The actual number of positive real roots may be the maximum, or the maximum decreased by a multiple of two.
• The maximum number of negative real roots can be found by counting the number of sign changes in f(-x). The actual number of negative real roots may be the maximum, or the maximum decreased by a multiple of two.
• Complex roots always come in pairs. That's why the number of positive or number of negative roots must decrease by two.

Consider: f(x) = 3x^6 + x^5 - x^4 + 3x^3 + 2x^2 - x + 1.
The signs in f(x) are + + - + + - +. There are 4 sign changes (+ to -) or (- to +).

Now, f(-x) = 3x^6 - x^5 - x^4 - 3x^3 + 2x^2 + x + 1.
The signs in f(-x) are + - - - + + +. There are 2 sign changes (+ to -) or (- to +).

Here are the Possible Combinations of Roots

Total	Positive	Negative	Complex
6	4	2	0
6	2	2	2
6	0	2	4
6	4	0	2
6	2	0	4
6	0	0	6

Notice that the positive and negative values can decrease by two independently of each other.

Rational Root Test

If a polynomial function has integer coefficients, then every rational zero will have the form p/q where p is a factor of the constant and q is a factor of the leading coefficient.

• Make sure the polynomial has integer coefficients. Multiply to get rid of fractions or decimals if need be (be sure to later divide).
• This only addresses the rational zeros. There may be real, but non-rational roots. There may be complex roots involving i.
• This says that rational zeros will have this form. It does not say that everything that has this form is a rational zero. What it does give you is a list of possible rational zeros

Example: f(x) = 4x^5 - x^2 + 12

Possible rational zeros will be of the form (factor of 12) over (factor of 4). A division table can help you find all these values

	1	2	3	4	6	12
1	1	2	3	4	6	12
2	1 / 2	1	3 / 2	2	3	6
4	1 / 4	1 / 2	3 / 4	1	3 / 2	3

The division table helps identify the possible rational zeros. You should throw out the duplicates, and list the others in order.

-12, -6, -4, -3, -2, -3 / 2, -1, -3 / 4, -1 / 2, -1 / 4, 1 / 4, 3 / 4, 1, 3 / 2, 2, 3, 4, 6, 12

Now, you perform synthetic division on possible rational zeros until you find one.

Here's where Descartes' Rule of Signs comes in. In this particular problem, there would be a maximum of 2 positive and 1 negative root. That means that you may have 2 or 0 positive roots, but you will always have 1 negative. There is no guarantee that negative is rational, though. Descartes only guaranteed real roots. If there were no negatives, then you would know not to try any.

Upper and Lower Bounds

If you have a polynomial with real coefficients and a positive leading coefficient, then ...

Upper Bound

If synthetic division is performed by dividing by x-k, where k>0, and all the signs in the bottom row of the synthetic division are non-negative, then x=k is an upper bound (nothing is larger) for the zeros of the polynomial.

Lower Bound

If synthetic division is performed by dividing by x-k, where k<0, and the signs in the bottom row of the synthetic division alternate (between non-negative and non-positive), then x=k is a lower bound (nothing is smaller) for the zeros of the polynomial.

The zero in the bottom row may be considered positive or negative as needed.

Suggested Attack to Finding Zeros of a Polynomial

1. Identify the total number of real or complex zeros (corollary to Fundamental Theorem of Algebra).
2. Identify the possible number of positive, negative, and complex zeros (Descartes' Rule of Signs).
3. List the possible rational zeros (Rational Root Theorem)
4. Try possible rational zeros until you find one that works. After each division by a positive value, check for possible upper bounds. After each division by a negative value, check for possible lower bounds (Upper and Lower Bound Theorems)
5. After you find a possible rational root that actually works, take the quotient and continue to try to factor it until it is down to a quadratic or less. Once it is a quadratic or less, there are other ways to solve it.
6. Write the linear and or linear / irreducible quadratic factorization (next section)

Really Important (and frustrating if you forget)!

Once you have found a zero using synthetic division, use the quotient as a new polynomial for all further divisions. The quotient will be one less degree than the original dividend. Each time you find a root, the quotient becomes one less in degree. Eventually, it will become a quadratic, and then you can factor, extract roots, complete the square, or use the quadratic equation to find the remaining roots.

If you continue to use the original function, you will become very frustrated and waste a lot of time.

Descartes' Sign Rule

A method of determining the maximum number of positive and negative real roots of a polynomial.

For positive roots, start with the sign of the coefficient of the lowest (or highest) power. Count the number of sign changes  as you proceed from the lowest to the highest power (ignoring powers which do not appear). Then  is the maximum number of positive roots. Furthermore, the number of allowable roots is , , , .... For example, consider the polynomial

(1)

Since there are three sign changes, there are a maximum of three possible positive roots.

For negative roots, starting with a polynomial , write a new polynomial  with the signs of all odd powers reversed, while leaving the signs of the even powers unchanged. Then proceed as before to count the number of sign changes . Then  is the maximum number of negative roots. For example, consider the polynomial

(2)

and compute the new polynomial

(3)

In this example, there are four sign changes, so there are a maximum of four negative roots.

Intermediate Value Theorem

The idea behind the Intermediate Value Theorem is this:

When you have two points connected by a continuous curve: one point below the line the other point above the line ... then there will be at least one place where the curve crosses the line!

Well of course you must cross the line to get from A to B!

Now that you know the idea, let's look more closely at the details.

Continuous

The curve must be continuous ... no gaps or jumps in it.

"Continuous" is a special term with an exact definition in calculus, but here we will use this simplified definition:

you can draw it without lifting your pen from the paper

More Formal

Here is that idea stated more formally:

		When: The curve is the function y = f(x), which is continuous on the interval [a, b], and w is a number between f(a) and f(b),
		Then ...

... there must be at least one value c within [a, b] such that f(c) = w

In other words the function y = f(x) at some point must be w = f(c)

Notice that:

• w is between f(a) and f(b), which leads to ...
• c must be between a and b

At Least One It also says "at least one value c", which means you could have more. Here, for example, are 3 points where f(x)=w.

How Polynomials Behave

A polynomial looks like this:

example of a polynomial

Continuous and Smooth

There are two main things about the graphs of Polynomials:

The graphs of polynomials are continuous, which is a special term with an exact definition in calculus, but here we will use this simplified definition:

you can draw it without lifting your pen from the paper

The graphs of polynomials are also smooth. No sharp "corners" or "cusps"

How the Curves Behave

Let us graph some polynomials to see what happens ...

... and let us start with the simplest form:

f(x) = xⁿ

Which actually does interesting things:

Even values of "n" behave the same: Always above (or equal to) 0 Always go through (0,0), (1,1) and (-1,1) Larger values of n flatten out near 0, and rise more sharply

And:

Odd values of "n" behave the same Always go from negative x and y to positive x and y Always go through (0,0), (1,1) and (-1,-1) Larger values of n flatten out near 0, and fall/rise more sharply

Power Function of Degree n

Next, by including a multiplier of a we get what is called a "Power Function":

f(x) = axⁿ
f(x) equals a times x to the "power" (ie exponent) n

The "a" changes it this way:

• Larger values of a squash the curve (inwards to y-axis)
• Smaller values of a expand it (away from y-axis)
• And negative values of a flip it upside down

Example: f(x) = ax2 a = 2, 1, ½, -1	Example: f(x) = ax3 a = 2, 1, ½, -1

We can use that knowledge when sketching some polynomials:

Example: Make a Sketch of y=1-2x⁷

Start with the simplest "odd power" graph of x³, and gradually turn it into 1-2x⁷

• You know how x³ looks,
• x⁷ will be similar, but flatter near zero, and steeper elsewhere,
• Squash it to get 2x⁷,
• Flip it to get -2x⁷, and
• Raise it by 1 to get 1-2x⁷.

Like this:

So by doing this step-by-step we can get a good result.

 

Turning Points

A Turning Point is an x-value where a local maximum or local minimum happens:

How many turning points does a polynomial have?

Never more than the Degree minus 1

The Degree of a Polynomial with one variable is the largest exponent of that variable.

Example: a polynomial of Degree 4 will have 3 turning points or less

x4-2x2+xhas 3 turning points		x4-2xhas only 1 turning point

The most is 3, but there can be less.

You may not know where they are, but at least you know the most there can be!

What Happens at the Ends

And when you move far from zero:

• far to the right (large values of x), or
• far to the left (large negative values of x)

then the graph starts to resemble the graph of y = axⁿ where axⁿ is the term with the highest degree.

Example: f(x) = 3x³-4x²+x

Far to the left or right, the graph will look like 3x³

Near Zero, they are  different		Far From Zero, they  become similar

This makes sense, because when x is large, then x³ is much greater than x² etc

This is officially called the "End Behavior Model".

And yes, we have come to the end!

Summary

Graphs will be continuous and smooth
Even exponents behave the same: above (or equal to) 0; go through (0,0), (1,1) and (-1,1); larger values of n flatten out near 0, and rise more sharply.
Odd exponents behave the same: go from negative x and y to positive x and y; go through (0,0), (1,1) and (-1,-1); larger values of n flatten out near 0, and fall/rise more sharply
Factors:
• Larger values squash the curve (inwards to y-axis)
• Smaller values expand it (away from y-axis)
• And negative values flip it upside down
Turning points: there will be "Degree-1" or less.
End Behavior: use the term with the largest exponent
Factors and Roots of a Polynomial Equation
Here are three important theorems relating to the roots of a polynomial:
(a) A polynomial of n-th degree can be factored into n linear factors.
(b) A polynomial equation of degree n has exactly n roots.
(c) If (x−r) is a factor of a polynomial, then x=r is a root of the associated polynomial equation.
Let's look at some examples to see what this means.
Example 1
The cubic polynomial f(x) = 4x³ − 3x² − 25x − 6 has degree 3 (since the highest power of x that appears is 3).
This polynomial can be factored (using Scientific Notebook or similar software) and written as
4x³ − 3x² − 25x − 6 = (x − 3)(4x + 1)(x + 2)
So we see that a 3rd degree polynomial has 3 roots.
The associated polynomial equation is formed by setting the polynomial equal to zero:
f(x) = 4x³ − 3x² − 25x − 6 = 0
In factored form, this is:
(x−3)(4x+1)(x+2)=0
We see from the expressions in brackets and using the 3rd theorem from above, that there are 3 roots, x=3,−14, −2.
In this example, all 3 roots of our polynomial equation of degree 3 are real.
Since (x−3) is a factor, then x=3 is a root.
Since (4x+1) is a factor, then x=−14 is a root.
Since (x+2) is a factor, then x=−2 is a root.
Example 2
The equationx⁵ − 4x⁴ − 7x³ + 14x² − 44x + 120 = 0 can be factored (using Scientific Notebook) and written as:
(x − 2)(x − 5)(x + 3)(x² + 4) = 0
We see there are 3 real roots x=2,5,−3, and 2complex roots x=±2j, (where j=−1−−−√).
So our 5th degree equation has 5 roots altogether.
[Do you need revision on complex numbers? Go toComplex Numbers.]
Example 3
In the previous section, Remainder Theorem and the Factor Theorem, we found in one of the examples that (x + 1) is a factor of f(x) = x³ + 2x² − 5x − 6.
This means that x=−1 is a root of x3+2x2−5x−6=0.
[To check this, substitute x=−1 into the polynomial. If it is a root, then you should get value 0 when you substitute.]
Example 4
The following polynomial equation would be rather tricky to solve using the Remainder and Factor Theorems. We will solve it using Scientific Notebook:
x⁴ + 0.4x³ − 6.49x² + 7.244x − 2.112 = 0
Answer
Note: Polynomial equations do not always have "nice" solutions! (By "nice solutions" I mean solutions which are integers or simple fractions.) This is why I feel the Remainder and Factor Theorems are pretty useless, because you can only use them if at least some of the solutions are integers or simple fractions.
If you use a computer algebra system (like Wolfram | Alpha or Scientific Notebook) to solve these, you can be done in seconds and move on to something more meaningful, like the applications.
Example 5
Solve the following polynomial equation using Scientific Notebook:
3x³ − x² − x + 4 = 0.
Answer
Using LiveMath to find Roots
We can also use LiveMath (or any similar Computer Algebra System) to find roots. This can be done graphically OR using the in-built root-finder.
Using a graph in LiveMath, we can easily find the roots of
x⁵ + 8.5x⁴ + 10x³ − 37.5x² − 36x + 54 = 0.
since the roots are the x-intercepts (i.e. where the function has value 0).
Let's see this how to do this in LiveMath:
LIVEMath
Complex Roots
Regarding complex roots, the following theorem applies :
If the coefficients of the equation f(x)=0 are real and a+bj is a complex root, then its conjugate a−bj is also a root.
For more on complex numbers, see: Complex Numbers
Example 6
In Example (2) above, we had 3 real roots and 2 complex roots. Those complex roots form a complex conjugate pair,
x = 0 − 2j and x = 0 + 2j
Example 7
The factors of the polynomial x³+ 7x² + 17x + 15 are found using LiveMath:
x³ + 7x² + 17x + 15 = (x + 3)(x + 2 − j)(x + 2 + j)
So the roots are
x=−3
x=−2+j andx=−2−j
There is one real root and the remaining 2 roots form a complex conjugate pair.
Remainder Theorem
and Factor Theorem
Or: how to avoid Polynomial Long Division when finding factors
Do you remember doing division in Arithmetic?

"7 divided by 2 equals 3 with a remainder of 1"
Each part of the division has names:

Which can be rewritten as a sum like this:

Polynomials
Well, we can also divide polynomials.
f(x) ÷ g(x) = q(x) with a remainder of r(x)
But it is better to write it as a sum like this:

I can show you this in an example using Polynomial Long Division:
Example: 2x²-5x-1 divided by x-3
• f(x) is 2x²-5x-1
• g(x) is x-3

After dividing we get the answer 2x+1, but there is a remainder of 2.
• q(x) is 2x+1
• r(x) is 2
In the style f(x) = g(x)·q(x) + r(x) we can write:
2x²-5x-1 = (x-3)(2x+1) + 2
But you need to know one more thing:
When you divide by a polynomial of degree 1 (such as "x-3") the remainder will have degree 0(in other words a constant, like "4").
And we are going to use that idea in the "Remainder Theorem":
The Remainder Theorem
When you divide a polynomial f(x) by x-c you get:
f(x) = (x-c)·q(x) + r(x)
But r(x) is simply the constant r (remember? when you divide by (x-c) the remainder is a constant) .... so we get this:
f(x) = (x-c)·q(x) + r
Now see what happens when you have x equal to c:
f(c) = (c-c)·q(c) + r
f(c) = (0)·q(c) + r
f(c) = r
So we get this:
The Remainder Theorem:
When you divide a polynomial f(x) by x-c the remainder r will be f(c)
So if you want to know the remainder after dividing by x-c you don't need to do any division:
Just calculate f(c).
Let us see that in practice:
Example: 2x²-5x-1 divided by x-3
(Continuing our example from above)
We don't need to divide by (x-3) ... just calculate f(3):
2(3)²-5(3)-1 = 2x9-5x3-1 = 18-15-1 = 2
And that is the remainder we got from our calculations above.
We didn't need to do Long Division at all!


Example: Dividing by x-4
(Continuing our examplee)


What would the remainder be if we divided by "x-4" ?
"c" is 4, so let us check f(4):
2(4)²-5(4)-1 = 2x16-5x4-1 = 32-20-1 = 11
Once again ... We didn't need to do Long Division to find it.
The Factor Theorem
Now ...
What if you calculated f(c) and it was 0?
... that means the remainder is 0, and ...
... (x-c) must be a factor of the polynomial!
Example: x²-3x-4
f(4) = (4)²-3(4)-4 = 16-12-4 = 0
so (x-4) must be a factor of x²-3x-4
And so we have:
The Factor Theorem:
When f(c)=0 then x-c is a factor of the polynomial
And the other way around, too:
When x-c is a factor of the polynomial then f(c)=0
Why Is This Useful?
Knowing that x-c is a factor is the same as knowing that c is a root (and vice versa).
The factor "x-c" and the root "c" are the same thing
Know one and you know the other
For one thing, it means that you can quickly check if (x-c) is a factor of the polynomial.
Example: 2x³-x²-7x+2
The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2. We can check easily:
f(2) = 2(2)³-(2)²-7(2)+2 = 16-4-14+2 = 0
Yes! f(2)=0, so we have found a root and a factor.


So (x-2) must be a factor of 2x³-x²-7x+2


How about where it crosses near -1.8?
f(-1.8) = 2(-1.8)³-(-1.8)²-7(-1.8)+2 = -11.664-3.24+12.6+2 = -0.304
No. (x+1.8) is not a factor.
Summary
The Remainder Theorem:
• When you divide a polynomial f(x) by x-c the remainder will be f(c)
The Factor Theorem:
• When f(c)=0 then x-c is a factor of the polynomial
• When x-c is a factor of the polynomial then f(c)=0

  The Rational Zeros Theorem

  Roots of a Polynomial

  A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0 .

  The Rational Zeros Theorem

  The Rational Zeros Theorem states:

  If P(x) is a polynomial with integer coefficients and if  is a zero of P(x) (P() = 0 ), then p is a factor of the constant term of P(x)and q is a factor of the leading coefficient of P(x) .

  We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps:

  1. Arrange the polynomial in descending order
  2. Write down all the factors of the constant term. These are all the possible values of p .
  3. Write down all the factors of the leading coefficient. These are all the possible values of q .
  4. Write down all the possible values of  . Remember that since factors can be negative,  and -  must both be included. Simplify each value and cross out any duplicates.
  5. Use synthetic division to determine the values of  for which P() = 0 . These are all the rational roots of P(x) .

  
  Example: Find all the rational zeros of P(x) = x ³ -9x + 9 + 2x ⁴ -19x ² .
  

  1. P(x) = 2x ⁴ + x ³ -19x ² - 9x + 9
  2. Factors of constant term: ±1 , ±3 , ±9 .
  3. Factors of leading coefficient: ±1 , ±2 .
  4. Possible values of  : ± , ± , ± , ± , ± , ± . These can be simplified to: ±1 , ± , ±3 , ± , ±9 , ± .
  5. Use synthetic division:

  

  Figure %: Synthetic Division

  Thus, the rational roots of P(x) are x = - 3 , -1 ,  , and 3 .

  We can often use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a . Next, we can use synthetic division to find one factor of the quotient. We can continue this process until the polynomial has been completely factored.

  
  Example (as above): Factor P(x) = 2x ⁴ + x ³ -19x ² - 9x + 9 .
  
  As seen from the second synthetic division above, 2x ⁴ + x ³ -19x ² -9x + 9÷x+ 1 = 2x ³ - x ² - 18x + 9 . Thus, P(x) = (x + 1)(2x ³ - x ² - 18x + 9) . The second term can be divided synthetically by x + 3 to yield 2x ² - 7x + 3 . Thus, P(x) = (x+ 1)(x + 3)(2x ² - 7x + 3) . The trinomial can then be factored into (x - 3)(2x - 1). Thus, P(x) = (x + 1)(x + 3)(x - 3)(2x - 1) . We can see that this solution is correct because the four rational roots found above are zeros of our result.