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
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$\,ax^k\,$, 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:
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$$
a_nx^n + a_{n1}x^{n1} + \cdots + a_1 + a_0
$$
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!

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 $\,x^2  2x + 1\,$ 
$(x1)^2 = 0$  factor: $\,x^2  2x + 1 = (x1)^2$ 
$x1 = 0$  $z^2 = 0$ 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 function—not just polynomials.
For example, the number $\,2\,$ is a zero of $\,f(x) = \sqrt{x  2}\,$ since $\,f(2) = \sqrt{22} = 0\,$.
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 function—they 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 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.
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 $\,x2\,$ is a factor of $\,P(x)\,$.
Why? Because it goes into $\,P(x)\,$ evenly; it goes in $\,x + 3\,$ times.
That is: $x^2 + x  6 = (x2)(x+3)\,$
Check by multiplying out: $(x2)(x+3) = x^2 + 3x  2x  6 = x^2 + x  6$
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 $\,P(x)\,$ $\,x4\,$ times, with a remainder of $\,14\,$.
That is: $\,x^2 + x  6 = (x+5)(x4) + 14$
Check by multiplying out and adding: $(x+5)(x4) + 14 = x^2  4x + 5x  20 + 14 = x^2 + x  6$
Note that when you write $\,P(x) = (x2)(x+3)\,$, the polynomial $\,P(x)\,$ has been expressed as a product (the last operation is multiplication).
However, when you write $\,P(x) = (x+5)(x4) + 14\,$, 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 $\,P(x) = (x2)(x+3)\,$, the zeros of $\,P\,$ jump out at you!
The number $\,2\,$ is a zero: $P(2) = (22)(2+3) = 0\cdot 5 = 0$
The number $\,3\,$ is a zero: $P(3) = (32)(3+3) = 5\cdot 0 = 0$
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:
