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.