For a particular type of function—a polynomial—we can say even more.
For a polynomial, there is a beautiful relationship between the zeroes and the factors.
For a polynomial $\,P\,$ and a real number $\,c\,$, the following are
equivalent:
Having said this, though, a study of the chart below will show that:
exactly one factor of $\,x  1\,$: $y = (x  1)^{\color{red}{1}} = x  1$ $1\,$ is a zero of multiplicity $\,\color{red}{1}$; that is, $1\,$ is a simple zero 
two factors of $\,x  1\,$: $y = (x  1)^{\color{red}{2}}$ $1\,$ is a zero of multiplicity $\,\color{red}{2}$ 
three factors of $\,x  1\,$: $y = (x  1)^{\color{red}{3}}$ $1\,$ is a zero of multiplicity $\,\color{red}{3}$ 
four factors of $\,x  1\,$: $y = (x  1)^{\color{red}{4}}$ $1\,$ is a zero of multiplicity $\,\color{red}{4}$ 
five factors of $\,x  1\,$: $y = (x  1)^{\color{red}{5}}$ $1\,$ is a zero of multiplicity $\,\color{red}{5}$ 
Let $\ P(x) = (x+2)x^5(x1)^3(x4)^2\,$. Then:
It is impossible to get a single toscale graph of $\,P\,$ that clearly shows the interesting behavior near all the zeroes. (The next section discusses this in more detail.)
Study the graphs at right and below, paying close attention to the scales on the $x$ and $y$axes: 
(A) 
(B) simple zero behavior at the zero: graph cuts straight through (no flattening) 
(C) zeros of ODD multiplicity greater than $1$ behavior at the zeroes: horizontal tangent lines; sign changes; the higher the multiplicity, the more flattening 
(D) zero of EVEN multiplicity behavior at the zero: horizontal tangent line; no sign change 
Let $\,P(x) = \frac 35(3x  5)^7(x)^4(8  x)^3\,$.
List the zeros of $\,P\,$, together with their multiplicities.
SOLUTION:
Since
$\ \cssId{s83}{(3x5)^7}
\cssId{s84}{= [ 3(x  \frac{5}{3}) ]^7}
\cssId{s85}{= 3^7(x  \frac{5}{3})^7}\ $,
the zero $\,\frac{5}{3}\,$ has multiplicity $\,7\,$.
You don't need to go through all this work:
just solve $\,3x  5 = 0\,$ to get the zero;
assign the power as the multiplicity.
Since $\,(x)^4 = (1)^4x^4 = x^4\ $,
the zero $\,0\,$ has multiplicity $\,4\,$.
Here's the shortcut:
solve ‘$\,x = 0\,$’ to get the zero;
assign the power as the multiplicity.
Since
$\,\cssId{s96}{(8  x)^3}
\cssId{s97}{= [1(x  8)]^3}
\cssId{s98}{= (1)^3(x8)^3}\,$,
the zero $\,8\,$ has multiplicity $\,3\,$.
The shortcut:
solve ‘$\,8  x = 0\,$’ to get the zero;
assign the power as the multiplicity.
zero  multiplicity 
$\frac{5}{3}$  $7$ 
$0$  $4$ 
$8$  $3$ 
multiplicity of zero  sign change at zero?  horizontal tangent line at zero? 
flattening behavior at zero 
sample graphs  

multiplicity $\,1\,$: a simple zero 
YES  NO 
no flattening; graph cuts straight through 


EVEN multiplicities $2$, $4$, $6$, $\ldots$ 
NO  YES  more flattening as multiplicity increases 


ODD multiplicities $3$, $5$, $7$, $\ldots$ 
YES  YES  more flattening as multiplicity increases 

Suppose that $\,c\,$ is a zero of a polynomial, so there is an $x$intercept at $\,c\,$.
Suppose you want to figure out if the sign of the polynomial changes (or not) as you pass through $\,c\,$.
Maybe it does change sign at $\,c\,$: perhaps it is negative on the left (graph below the $x$axis) and positive on the right (graph above the $x$axis). 

Or, maybe it doesn't change sign at $\,c\,$: perhaps it is negative both to the left and right of $\,c\,$. 
Here's the good news:
Consider the same polynomial as above,
$\ P(x) = (x+2)x^5(x1)^3(x4)^2\,$.
Let's decide if there's a sign change at $\,x = 1\,$.
Write the function as
$\,\cssId{s150}{P(x) = (x1)^3\ \cdot\ }
\cssId{s151}{\overbrace{x^5(x+2)(x4)^2}^{\text{stuff}}}\,$.
Notice that all the factors of $\,(x1)\,$ have been pulled out to the front,
and the remaining factors have been labeled ‘stuff’.
The number line below shows all the zeroes of $\,P\,$.
Recall that zeroes
are the only type of place where a polynomial can change its sign,
since there are no breaks in its graph.
The interval highlighted in yellow doesn't contain any zero except the one under consideration. Since a polynomial has a finite number of zeroes, you can always get such an interval. For example, you can always go halfway to the closest zero on the left, and halfway to the closest zero on the right, as was done here.
Inside the yellow interval, ‘stuff’ has a constant sign:
Since ‘stuff’ doesn't change its sign inside the yellow interval, the only part of the formula that affects the sign of $\,P(x)\,$ in the yellow interval is $\,(x1)^3\,$.
To the left of $\,1\,$, $\,(x1)^3\,$ is negative.
To the right of $\,1\,$, $\,(x1)^3\,$ is positive.
So, $\,P(x)\,$ does change sign at $\,1\,$.
Let $\,P(x) = 11(x+7)^5(x+3)^8(x+1)(x^4)(x  5)^2(x  8)^3\,$.
The zeroes of $\,P\,$ are:
$7,\ 3,\ 1,\ 0,\ 5,\ 8\,$
The function $\,P\,$ changes its sign at $7$, $1$, and $8$,
since the corresponding factors are raised to odd powers.
The function $\,P\,$ does not change its sign at $3$, $0$, and $5$,
since the corresponding factors are raised to even powers.
A small number (between $\,1\,$ and $\,1\,$),
when multiplied by itself,
gets even smaller (closer to zero).
The number $\,x  c\,$ is small when $\,x\,$ is close to $\,c\,$.
So, when $\,x  c\,$ gets repeatedly multiplied by itself,
it gets smaller, and smaller, and smaller.
Look at the numbers and graphs below—for higher multiplicities, the outputs are smaller, so the points are closer to the $x$axis!
$x$  $(x  3)$  $(x  3)^2$  $(x  3)^3$  $(x  3)^4$  $(x  3)^5$ 
$3.1$  $0.1$  $0.01$  $0.001$  $0.0001$  $0.00001$ 
all these graphs have the same horizontal and vertical scales 
$y = x  3$ point $(3.1\,,\,0.1)$ 
$y = (x  3)^2$ point $(3.1\,,\,0.01)$ 
$y = (x  3)^3$ point $(3.1\,,\,0.001)$ 
$y = (x  3)^4$ point $(3.1\,,\,0.0001)$ 
$y = (x  3)^5$ point $(3.1\,,\,0.00001)$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
