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:
the definition of a polynomial |
A 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, $\,k\in \{0,1,2,3,\ldots\}\,$. |
how to put a polynomial in standard form |
The 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. |
how to determine a polynomial's degree | The degree of the polynomial is the highest power to which $\,x\,$ is raised. |
how to determine a polynomial's leading coefficient |
The coefficient of the highest power term
is called the leading coefficient of the polynomial. When a polynomial is written in standard form, the leading coefficient actually leads the polynomial. |
For example, $\,P(x) = -3x + 11 - 5x^7 + 9x^2\,$ is a polynomial.
Its standard form is: $\,P(x) = -5x^7 + 9x^2 - 3x + 11$
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.
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.
It would be reasonable to guess that a zero of a function has something to do with the number $\,0\,$.
It does!
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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. |
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 |
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 |
You can talk about zeros of any functionnot 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.
There are several equivalent ways to think about zeros of functions:
In particular, observe that zeros are easy to spot if you have the graph of a functionthey are the $\,x\,$-intercepts.
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 remainderit 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.
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.)
For example, let $\,P(x) = x^2 + x - 6\,$.
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:
$(x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6$
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:
$\,x^2 + x - 6 = (x+5)(x-4) + 14$
Check by multiplying out and adding:
$(x+5)(x-4) + 14 = x^2 - 4x + 5x - 20 + 14 = x^2 + x - 6$
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.
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:
$P(2) = (2-2)(2+3) = 0\cdot 5 = 0$
The number $\,-3\,$ is a zero:
$P(-3) = (-3-2)(-3+3) = -5\cdot 0 = 0$
Returning to the other name (just for fun):
$P(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$
$P(-3) = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0$
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!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
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