# Introduction to Polynomials

First, review the Introduction to Polynomials lesson in the Algebra II curriculum to make sure you've mastered basic polynomial concepts and terminology.

For example, you should know:

*polynomial*is a finite sum of terms, each of the form $\,ax^k\,,$ where $\,a\,$ is a real number, and $\,k\,$ is a nonnegative integer. That is: $\,k\in \{0,1,2,3,\ldots\}$

*standard form of a polynomial*is: $$ \cssId{s11}{a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 + a_0} $$ Here, $\,n\,$ denotes the highest power to which $\,x\,$ is raised.

Thus, in standard form, the highest power term is listed first, and subsequent powers are listed in decreasing order.

*degree*of the polynomial is the highest power to which $\,x\,$ is raised.

*leading coefficient*of the polynomial.

When a polynomial is written in standard form,
the leading coefficient actually
*leads* the polynomial.

Its degree is $\,7\,.$ Its leading coefficient is $\,-5\,.$

For a more thorough review and practice exercises, study Introduction to Polynomials in the Algebra II curriculum.

## Relationship between the Zeros and Factors of a Polynomial

There is a beautiful relationship
between the zeros and
factors of a polynomial,
which is explored in the next section.
In preparation,
the concepts of *zero* and *factor*
are reviewed below.

## Zeros (Roots) of Functions

It would be reasonable to guess
that a *zero of a function*
has something to do with the number $\,0\,.$
It does!

What input (or inputs) must be dropped in the top of a function box, to get zero out the bottom?

A *zero of a function*
is an *input*
whose corresponding output
is zero.

## Example (Finding the Zeros of a Function)

Consider the function $\,f(x) = x^2 - 2x + 1\,.$ Recall that $\,f(x)\,$ represents the output from the function $\,f\,$ when the input is $\,x\,.$

What value(s) of $\,x\,$ make $\,f(x)\,$ equal to $\,0\,$? | |

$f(x) = 0$ | Set the output, $\,f(x)\,,$ equal to zero. |

$x^2 - 2x + 1 = 0$ | Substitution: $\,f(x)\,$ is $\,x^2 - 2x + 1\,$ |

$(x-1)^2 = 0$ | Factor: $\,x^2 - 2x + 1 = (x-1)^2$ |

$x-1 = 0$ | $z^2 = 0\,$ if and only if $\,z = 0$ |

$x = 1$ | Add $\,1\,$ to both sides. |

In this example, there is only one input, $\,x = 1\,,$ for which the output is zero.

Check:

$$\cssId{s54}{f(1) = 1^2 - 2(1) + 1 = 0}$$Here, $\,1\,$ is the only zero of the function $\,f\,.$

A zero of a function is also called a *root* of the function.

Initially, you might find yourself a bit uncomfortable saying ‘$\,1\,$ is a zero’. It almost sounds like you're saying ‘$\,1 = 0\,$’, which is of course false. The word ‘a’ is critically important in the sentence!

When you say: | $\,1\,$ is a zero |

You're saying: | $\,1\,$ is an input whose corresponding output is zero |

## Example (Zeros of Non-Polynomials)

You can talk about zeros of *any* function—not just polynomials.

For example, the number $\,2\,$ is a zero of $\,f(x) = \sqrt{x - 2}\,$ since $\,f(2) = \sqrt{2-2} = 0\,.$

The function $\,f(x) = {\text{e}}^x\,$ doesn't have any zeros, since $\,{\text{e}}^x\,$ is always strictly greater than zero.

## Equivalent Characterizations of a Zero of a Function

There are several equivalent ways to think about zeros of functions:

A *zero* (or *root*) of a function
is an input, whose corresponding output is zero.

Let $\,f\,$ be a function, and let $\,c\,$ be an input to $\,f\,.$

Then, the following are equivalent:

- $c\,$ is a zero of $\,f$
- $c\,$ is a root of $\,f$
- $c\,$ is an input, with corresponding output $\,0$
- $f(c) = 0$
- the point $\,(c,0)\,$ lies on the graph of $\,f\,$
- $f\,$ has an $x$-intercept equal to $\,c$
- the graph of $\,f\,$ crosses the $x$-axis at $\,c\,$
- $x = c\,$ is a solution of the equation $\,f(x) = 0$

In particular, observe that zeros are easy to spot if you have the graph of a function: they are the $x$-intercepts.

## Review: Factors of a (Whole) Number

Early on, you learned that the factors of (say) $\,27\,$ are numbers that go into $\,27\,$ evenly (with no remainder). ‘No remainder’ means the same thing as a remainder of zero.

For example, $\,3\,$ is a factor of $\,27\,.$ Why? Because it goes into $\,27\,$ nine times: $27 = 3\cdot 9$

The number (say) $\,11\,$ is
*not* a factor of $\,27\,.$
Why not?
Because there is a nonzero remainder—it goes in twice, with a remainder of $\,5\,.$
That is: $\,27 = 2\cdot 11 + 5$

Note that when you write $\,27 = 3\cdot 9\,,$ the number $\,27\,$ has been expressed as a product (the last operation is multiplication). However, when you write $\,27 = 2\cdot 11 + 5\,,$ the number $\,27\,$ has been expressed as a sum (the last operation is addition).

Key Idea:
The *factors* of a number can be used
to express the number as a product.
Indeed, *factoring* means
to take something and express it as a product.

## Factors of a Polynomial

In a similar way, we can talk about factors of a polynomial. These are expressions that go into the polynomial evenly (with no remainder). (Note: long division of polynomials is covered in a future section.)

The expression $\,x-2\,$ is a factor of $\,P(x)\,.$ Why? Because it goes into $\,P(x)\,$ evenly: it goes in $\,x + 3\,$ times.

That is: $$x^2 + x - 6 = (x-2)(x+3)$$

Check by multiplying out:

The expression (say) $\,x+5\,$
is *not* a factor of $\,P(x)\,.$
Why not?
Because when $\,x+5\,$ is divided
into $\,P(x)\,,$ there is a nonzero remainder.

Indeed, long division of polynomials shows that $\,x+5\,$ goes into $\,P(x)\,$ $\,x-4\,$ times, with a remainder of $\,14\,.$ That is:

$$\cssId{s124}{x^2 + x - 6 = (x+5)(x-4) + 14}$$Check by multiplying out and adding:

Note that when you write $\,P(x) = (x-2)(x+3)\,,$ the polynomial $\,P(x)\,$ has been expressed as a product (the last operation is multiplication).

However, when you write $\,P(x) = (x+5)(x-4) + 14\,,$ the polynomial $\,P(x)\,$ has been expressed as a sum (the last operation is addition).

Key Idea:
The *factors* of a polynomial
can be used to express the polynomial as a product.

## Looking Ahead to the Next Section...

Expressions have lots of different names, and different names are good for different things.

When you write $\,P(x) = x^2 + x - 6\,,$ it's not the least bit clear (at least to this author) what value(s) of $\,x\,$ make the output zero. What number, when squared, then added to itself, then with $\,6\,$ subtracted, gives zero? Not a clue!

However, when you write $\,P(x) = (x-2)(x+3)\,,$ the zeros of $\,P\,$ jump out at you!

The number $\,2\,$ is a zero:

$$ \begin{align} P(2) &= (2-2)(2+3)\cr &= 0\cdot 5\cr &= 0 \end{align} $$

The number $\,-3\,$ is a zero:

$$ \begin{align} P(-3) &= (-3-2)(-3+3)\cr &= -5\cdot 0\cr &= 0 \end{align} $$

Returning to the other name (just for fun):

$$ \cssId{s146}{ \begin{align} P(2) &= 2^2 + 2 - 6\cr &= 4 + 2 - 6\cr &= 0 \end{align}} $$ $$ \cssId{s147}{ \begin{align} P(-3) &= (-3)^2 + (-3) - 6\cr &= 9 - 3 - 6\cr &= 0 \end{align}} $$You may have noticed the beautiful relationship between zeros and factors:

$x-\color{blue}{2}$ is a factor;

$\color{blue}{2}\,$ is a zero

$x + 3 = x-(\color{blue}{-3})$ is a factor;

$\color{blue}{-3}\,$ is a zero

This observation is formalized in the next section!