Question 369474
<pre>
I need help with sketching the graph of this factored polynomial 
{{{y=(x+1)^4 (x-2)^3 (x-1)}}}

The roots, zeros, x-intercepts, or solutions 

(they are called by all four names at different times and are the
places where the curve comes in contact with the x-axis) 

are found by setting the expressions in the parentheses = 0

x+1=0, x-2=0, x-1=0

and solving, getting

x=-1, x=2, and x=1

When you have something like this:  {{{(x-R)^N}}} as a factor
of a polynomial then we say R is a zero with multiplicity of N.  
If N is an even number the graphed curve "bounces" off the x-axis 
at R, and if N  is an odd number, the graphed curve cuts through 
the x-axis at R.

Since the exponent 4 in {{{(x+1)^4}}} is even, we know that the 
graphed curve "bounces off" the x-axis at -1.

Since the exponent 3 in {{{(x-2)^3}}} is odd, we know that the 
graphed curve cuts through the x-axis at 2.

Since the understood exponent 1 in {{{(x-1)}}} is odd, we know that the 
graphed curve cuts through the x-axis at 1.

The y-intercept is found by substituting 0 for x in

{{{y=(0+1)^4 (0-2)^3 (0-1)}}}

{{{y=(1)^4(-2)^3 (-1)}}}

{{{y=(1)(-8)(-1)}}}

{{{y=8}}}

So we draw the graph this way.  We start at the y-intercept and draw
downward to the left so that the curve bounces off the x-axis at -1.
Then to the right it eventually goes down and cuts through the x-axis
at 1 and 2. 

Here is the graph

{{{graph(4000/17,800,-2,3,-5,12,(x+1)^4 (x-2)^3 (x-1))}}}

The graph cuts through the x-axis at 1 and 2 and "bounces off" the 
x-axis at -1.  To draw it accurately like this you would
have to plot some more points.

Edwin</pre>